Math E-learning Materials - Lyskalyska.net/MathProject/Projects/Math e-learning/Math E-learing Materi… · Math E-learning Materials Gymnazium Teplice, Teplice, Czech Republic Lyseopuistonlukio,

This project has been funded with support from the European Commission.

This publication reflects the views only of the author, and the Commission cannot be held

responsible for any use which may be made of the information contained therein.

Math E-learning Materials

Gymnazium Teplice, Teplice, Czech RepublicLyseopuistonlukio, Rovaniemi, Finland

The ZST Technical School, Mikolow, PolandGymnazium A. Bernolaka, Namestovo, Slovakia

Tuesday, May 4, 2010 1. M6 - Math E-learing Materials FIN, PL, CZ, SLOVAKIA.nb

Site: www.deltasoft.at M@th Desktop 1

Description of the ProjectOpen êClose Print

The main aim of our project „Math E-learning Materials“ is to prepare e-learning

materials which could be used at schools both by teachers and students. Producedteaching units and materials can be used as an enrichment in order to provide engaginginteractive lessons blended with standard teaching methods. We would like to encourageteachers to use blended learning methods. The produced materials can be found on theproject homepage, so it would be easily accessible also to students who can use them forsome extra practising.

Although our schools have different curricula, we have found what we have incommon. So we had decided to cover various topics (Algebra, Functions, Trigonometryand Probability). The produced notebooks contain some theory, examples and graphs.

Students had a chance to work together and to get to know each other during localproject meetings in Mikolow - Poland, Teplice - Czech Republic, Graz - Austria andRovaniemi - Finland. This cooperation continued via emails, Skype and videoconferences.They did work together on tasks given by the teachers, they exchanged their solutionsand discussed the problems.

AlgebraOpen êClose Print

Linear equations and applicationsOpen ê Close

1. Find four consecutive even integers such that the sum of the first three exceeds the fourth by 8.



Let x = the first even integer,then x, x+2, x+4 and x+6 represent four consecutive even integers starting with the even integer x.The phrase "the sum of the first three exceeds the fourth by 8" translates into an equation: Sum of the first three = fourth + excess.x + (x+2) + (x+ 4) == (x + 6) + 8

Input @Æ MDRealOnly ;

Clear @xD;

Solve @8x + Hx + 2L + Hx + 4L � Hx + 6L + 8<, 8x< D88x → 4<<

Answer: The four consecutive integers are 4, 6, 8 and 10.

2. If one side of a triangle is one-third the perimeter, the second side is 7 meters, and the third side is one-fifth the perimeter, what is the perimeter of the triangle?

Input @

MDPlot B80<, 8x, 0, 7 <, PlotRange → 8−0.1, 7 <,

AspectRatio → 1, Epilog → :PointSize @0.025 D, Point @80, 0 <D,

Point @87, 0 <D, Point @82.5, 2 <D, Line @880, 0 <, 82.5, 2 <<D,

Line @882.5, 2 <, 87, 0 <<D, Text B"p

3" , 85, 1.4 <F,

Text B"p

5" , 81, 1.3 <F, Text @"7" , 83.5, 0.4 <D> F;



1 2 3 4 5 6 7x

1

2

3

4

5

6

7y

pÅÅÅÅÅÅ3

pÅÅÅÅÅÅ5

7

Let p = the perimeter.Perimetr of triangle is p = a + b + c.

Åp

3+ Å

p

5+ 7 = p

Input @

Æ MDRealOnly ;

Clear @pD;

Solve B: p

3+

p

5+ 7 � p>, 8p< F

88p → 15<<

Answer: The perimeter is 15 meters.

3.

The distance along a shipping route between San Francisco and Honolulu is 2100 nautical miles. If one ship leaves San Francisco at the same time another leaves Honolulu, and if the former travels at 15 knots and the latter at 20 knots, how long will it take the two ships to rendezvous? How far will they be from Honolulu and San Francisco at that time?



Let t = number of hours until both ships meetdistance = speed x timed1 = distance ship 1 from Honolulu travels to meeting point d1 = 20 td2 = distance ship 2 from San Francisco travels to meeting point d2 = 15 t2 100 = total distance from Honolulu to San Franciscod1 + d2 = 210020 t + 15 t = 2100


Clear @t D;

Solve @8 20 t + 15 t == 2100 <, 8t < D88t → 60<<

Input @ t = 60;

Input @ 20 t

1200

Input @ 15 t

900

Answer: It takes 60 hours for the ships to meet.Distance from Honolulu is 1200 nautical miles.Distance from San Francisco is 900 nautical miles.

4. An excursion boat takes 1.5 times as long to go 360 miles up a river than to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?



Let x = rate of current (in miles per hour)15 - x = rate of boat upstream15 + x = rate of boat downstreamtime upstream = (1,5) * (time downstream)

time = distance

rate

distance upstream

rate upstream= H1, 5L *

distance downstream

rate downstream

360

15 - x = 1,5 *

360

15 + x

Input @

Æ MDRealOnly ;

Clear @xD;

Solve B: 360

15 − x== H1.5 L

360

15 + x>, 8x< F

88x → 3. <<

Answer: The rate of the current is 3 miles per hour.

5. An advertising firm has an old computer that can prepare a whole mailing in 6 hours. With the help of a newer model the job is complete in 2 hours. How long would it take the newer model to do the job alone?



Let x = time (in hours) for the newer model to do the job alone(part of job completed in a given lingth of time) = (rate) * (time)

rate of old model = 1

6job per hour

rate of newer model = 1

xjob per hour

(part of job completed by old model in 2 hours) + (part of job completed by new model in 2 hours) = 1 whole job

(rate of old model) * (time of old model) + (rate of new model) * (time of new model) = 1 whole job

Å1

6* 2 + Å

1

x* 2 = 1

Input @

Æ MDRealOnly ;

Clear @xD;

Solve B: 1

6 2 +

1

x 2 == 1>, 8x < F

88x → 3<<

Answer: The new computer could do the job alone in 3 hours.



6. How many liters of a mixture containing 80% alcohol should be added to 5 liters of a 20% solution to yield a 30% solution?

Let x = amount of 80% solution used(amount of alcohol in first solution) + (amount of alcohol in second solution) = (amount of alcohol in mixture)

0,8 x + 0,2 * 5 = 0,3 * (x + 5)


Clear @xD;

Solve @80.8 x + 5 H0.2 L == 0.3 Hx + 5L<, 8x< D88x → 1. <<

Answer: Add 1 liter of the 80% solution.

Systems of linear equations and applicationsOpen ê Close

7.

An individual wants to use milk and orange juice to increase the amount of calcium and vitamin A in her daily diet. An ounce of milk contains 38 milligrams of calcium and 56 micrograms of vitamin A. An ounce of orange juice contains 5 milligrams of calcium and 60 micrograms of vitamin A. How many ounces of milk and orange juice should she drink each day to provide exactly 550 milligrams of calcium and 1 200 micrograms of vitamin A?



Let x = number of ounces of milk y = number of ounces of orange juice (calcium in x oz of milk) + (calcium in y oz of orange juice) = (total calcium needed (milligrams))38 x + 5 y = 550

(vitamin A in x oz of milk) + (vitamin A in y oz of orange juice) = (total vitamin A needed (micrograms))56 x + 60 y = 1 200


Clear @xD;

NSolve @838 x + 5 y == 550, 56 x + 60 y == 1200<, 8x, y < D88x → 13.5, y → 7.4 <<

Answer: Drinking 13.5 ounces of milk and 7.4 ounces of orange juice each day will provide the required amounts of calcium and vitamin A.

8.

An airplane makes the 2400 mile trip from Washington, D.C., to San Francisco in 7.5 hours and makes the return trip in 6 hours. Assuming that the plane travels at a constant airspeed and that the wind blows at a constant rate from west to east, find the plane's airspeed and the wind rate.



Let x = airspeed of the plane y = the rate at which the wind is blowingThe ground speed of the plane is determined by combining these two rates:x - y = ground speed flying east to west (headwind)x + y = ground speed flying west to east (tailwind)

The formula of distance: d = r * t H d = distance, r = rate and t = time)2 400 = 7.5 * (x - y) from Washington to San Francisco2 400 = 6 * (x + y) from San Francisco to Washington


Clear @x, y D;

Solve @82400 == 7.5 ∗ Hx − yL, 2400 == 6 ∗ Hx + yL<, 8x, y < D88x → 360., y → 40. <<

Answer: The plane's airspeed is 360 mph and the wind rate is 40 mph.

9.

A publisher is planning to produce a new text-book. The fixed-costs (reviewing, editing, typesetting, and so on) are $320 000, and the variable costs (printing, sales commissions, and so on) are $31.25 per book. The wholesale price (the amount received by the publisher) will be $43.75 per book. How many books must the publisher sell to break even; that is, so that costs will equal revenues?



Let x = the number of books printed and sold, then the cost and revenue equations for the publisher arey = 320 000 + 31.25 x cost equationy = 43.75 xThe publisher breaks even when costs equal revenues. Using the second equation to substitute for y in the first equation320 000 + 31.25 x ã 43.75 x


Clear @xD;

Solve @8320 000 + 31.25 x � 43.75 x<, 8x<D88x → 25 600. <<

Answer: The publisher wil break even when 25 600 books are printed and sold.

10.Suppose you have $12 000 to invest. If part is invested at 10% and the rest at 15%, how much should be invested at each rate to yield 12% on the total amount invested?



Let x = the first part of investment y = the second part of investment x + y = invested amount = 12 000 0.1 (the thirst part of investment) + 0.15 (the second part of investment) = 0.12 (invested amount) 0.1 x + 0.15 y = 0.12 (x + y) x + y = 12 000


Clear @x, y D;

NSolve @80.1 x + 0.15 y == 0.12 Hx + yL, x + y == 12 000 <, 8x, y < D88x → 7200., y → 4800. <<

Answer: At 10% rate, $7 200 should be invested. At 15% rate, $4 800 should be invested.

11.A chemist has two solutions of hydrochloric acid in stock: a 50% solution and an 80% solution. How much of each should be used to obtain 100 mililiters of a 68% solution?



Let x = amount of 50% solution y = amount of 80% solution x + y = 100 milliliters68% = the target solution0.5 x + 0.8 y = 0.68 (x + y)


Clear @x, y D;

NSolve @80.5 x + 0.8 y == 0.68 Hx + yL, x + y == 100<, 8x, y <D88x → 40., y → 60. <<

Answer: The chemist should use 40 ml of 50% solution and 60 ml of 80% solution.

12.

A fruit grower can use two types of fertilizer in an orange grove, brand A and brand B. Each bag of brand A contains 8 pounds of nitrogen and 4 pounds of phosphoric acid. Each bag of brand B contains 7 pounds of nitrogen and 7 pounds of phosphoric acid. Tests indicate that the grove needs 720 pounds of nitrogen and 500 pounds of phosphoric acid. How many bags of each brand should be used to provide the required amounts of nitrogen and phosphoric acid?



Let x = the number of bags of brand A y = the number of bags of brand B one bag of brand A = 8 pounds of nitrogen and 4 pounds of phosphoric acidone bag of brand B = 7 pounds of nitrogen and 7 pounds of phosphoric acidamount of nitrogen = 720 poundsamount of phosphoric acid = 500 pounds 8 x + 7 y = 7204 x + 7 y = 500


Clear @x, y D;

NSolve @88 x + 7 y == 720, 4 x + 7 y == 500<, 8x, y <D88x → 55., y → 40. <<

Answer: 55 bags of brand A fertilizer and 40 bags of brand B fertilizer should be used.



Quadratic equations and applicationsOpen ê Close

13. The sum of a number and its reciprocal is 13

6. Find all such numbers.

Let x = the number

Å1

x= the reciprocal number

(the number) + (the reciprocal number) = 13

6

x + Å1

x =

13

6

Input @

Æ MDRealOnly ;

Clear @xD;

Solve B:x +1

x==

13

6>, 8x < F

::x →2

3>, :x →

3

2>>

Answer: The numbers are Å2

3and Å

3

2.

14. An excursion boat takes 1.6 hours longer to go 36 km up a river than to return. If the rate of the current is 4 km per hour, what is the rate of the boat in still water?



Let x = rate of boat in still water x + 4 = rate downstream x - 4 = rate upstreamdistance = 36 km

the formula of time is distance

rate

(time upstream) - (time downstream) = 1.636

x - 4 -

36

x + 4 = 1,6

Input @

Clear @xD;

Solve B 36

x − 4−

36

x + 4== 1.6, 8x< F

88x → −14. <, 8x → 14. <<

Note: Because time can't have negative values, the first solution is excluded.

Answer: The rate in still water is 14 km/h.

15.

A payroll can be completed in 4 hours by two computers working simultaneously. How many hours are required for each computer to complete the payroll alone if the older model requires 3 hours longer than the newer model? Compute answers to two decimal places.



Let x = time for new model to complete the payroll alone x + 3 = time for old model to complete the payroll alone 4 = time for both computers to complete the payroll together1

x= rate for new model

1

x + 3= rate for old model

(part of job completed by new model in 4 hours) + (part of job completed by old model in 4 hours) = 1 whole job

Å1

x* 4 +

1

x + 3* 4 = 1

Input @

Æ MDRealOnly ;

Clear @xD;

Solve B: 1

x 4 +

1

x + 3 4 == 1>, 8x<F

::x →1

2J5 − 73 N>, :x →

1

2J5 + 73 N>>

Note: Because time can't have negative values, the first solution is excluded.

Input @ x =1

2J5 + 73 N êê N

6.772



Input @ x + 3

9.772

Answer: The new model would complete the payroll in 6.77 hours working alone, and the old model would complete the payroll in 9.77 hours working alone.

16. Find two numbers such that their sum is 21 and their product is 104.

Let x = the first number y = the second numbersum of numbers is x + y = 21 product of numbers is x * y = 104


Clear @x, y D;

Solve @8x + y == 21, x y == 104<, 8x, y < D88x → 8, y → 13<, 8x → 13, y → 8<<

Answer: The numbers are 8 and 13.

17. Find all numbers with the property that when the number is added to itself the sum is the same as when the number is multiplied by itself.



Let x = the number sum of numbers is x + x product of numbers is x * x x + x = x * x


Clear @xD;

Solve @8x + x == x x<, 8x< D88x → 0<, 8x → 2<<

Answer: The numbers are 0 and 2.

18. Find two consecutive positive even integers whose product is 168.

Let x = the first integer x + 2 = the second integerproduct of integers is 168x * (x + 2) = 168


Clear @xD;

Solve @8x Hx + 2L == 168<, 8x<D88x → −14<, 8x → 12<<

Note: Only positive even integers.

Input @ x = 12

12

Input @ x + 2

14

Answer: The integers are 12 and 14.



If the length and width of a 4 m by 2 m rectangle are each increased by the same amount, the area of the new rectangle will be twice that of the original. What are the dimensions of the new rectangle (to two decimal places)?

Input @

MDPlot A80<, 8x, 0, 5.12 <, PlotRange → 80, 3.12 <,

Epilog → 9Purple, Line @880, 3.12 <, 85.12, 3.12 <<D,

Text @"4 + x" , 82.5, 0.3 <D, Line @885.12, 0 <, 85.12, 3.12 <<D,

Line @880, 12 <, 815, 12 <<D, Text A"S = H4 ∗ 2L∗2 m2" ,

82.5, 1.5 <E, Text @"2 + x" , 80.5, 2 <D=E;

1 2 3 4 5x

0.5

1

1.5

2

2.5

3

y

4 + x

S = H4 * 2L*2 m2

2 + x

Let 4 + x = the length of the new rectangle 2 + x = the width of the new rectangle2 * 4 area of the rectangle(2 * 4) * 2 area of the new rectangle(x + 4) * (x + 2) = (2 * 4) * 2


Clear @xD;

Solve @8Hx + 4L Hx + 2L == 2 × 4 × 2<, 8x< D

::x → −3 − 17 >, :x → −3 + 17 >>

Note: Becasue the length and the width can't have negative values, the first solution is excluded.

Input @ x = −3 + 17 êê N

1.12311

Input @ x + 4



5.12311

Input @ x + 2

3.12311

Answer: The dimensions are 5.12 m and 3.12 m.

20.Find the base b and height h of a triangle with an area of 2 square m if its base

is 3 m longer than its height and the formula for area is S = Å1

2 b.h.

Input @

MDPlot A80<, 8x, 0, 4 <, PlotRange → 8−0.1, 2 <,

Epilog → 9Purple, Line @880, 1 <, 84, 0 <<D,

Text @"h" , 80.2, 0.5 <D, Text @"b + 3" , 81.5, 0.1 <D,

Text A"S = 2m2" , 81, 0.4 <E, PointSize @0.025 D,

Point @84, 0 <D, Point @80, 0 <D, Point @80, 1 <D=E;

1 2 3 4x

0.5

1

1.5

2y

h

b + 3

S= 2m2

Let height h base b = h + 3 area S = 2

area S = Å1

2 b * h

Å1

2* (h + 3) * h = 2



Input @

Æ MDRealOnly ;

Clear @hD;

Solve B: 1

2 Hh + 3L h == 2>, 8h < F

88h → −4<, 8h → 1<<

Note: The height can't have negative values.

Input @ h = 1

1

Input @ h + 3

4

Answer: The base is 4 m and the height is 1 m.

21.Two planes travel at right angles to each other after leaving the same airport at the same time. One hour later they are 260 km apart. If one travels 140 km per hour faster than the other, what is the rate of each?

Input @

MDPlot @80<, 8x, 0, 250 <, PlotRange → 8−2, 102 <,


Text @"260 km" , 8130, 60 <D, Text @"x" , 820, 50 <D,

Text @"x + 140" , 8100, 10 <D, PointSize @0.025 D,

Point @80, 0 <D, Point @8240, 0 <D, Point @80, 100 <D<D;

50 100 150 200 250x

20

40

60

80

100

y

260 kmx

x + 140

Input @ 2



Let x = rate of the first planex + 140 = rate of the second planedistance = 260 km

Pythagorean theorem x2 + Hx + 140L2 = 2602

Input @

Æ MDRealOnly ;

Clear @xD;

Solve A9 x2 + Hx + 140L2 == 260 2 =, 8x < E88x → −240<, 8x → 100<<

Note: The rate can't have negative values.

Input @ x = 100

100

Input @ x + 140

240

Answer: The rates of the planes are 100 km/h and 240 km/h.

22.One pipe can fill a tank in 5 hours less than another. Together they can fill the tank in 5 hours. How long would it take to each alone to fill the tank? Compute the answer to two decimal places.

Let x = time for first pipe to fill the tank alone x + 5 = time for second pipe to fill the tank alone 5 = time for both pipes to fill the tank together



Å1

x* 5 +

1

x + 5* 5 = 1

Input @

Æ MDRealOnly ;

Clear @xD;

Solve B: 1

x 5 +

1

x + 5 5 == 1 >, 8 x < F

::x →5

2J1 − 5 N>, :x →

5

2J1 + 5 N>>

Note: The time can't have negative values.

Input @ x =5

2J1 + 5 N êê N

8.09017

Input @ x + 5

13.0902

Answer: The faster pipe could fill the tank in 8.09 hours. The slower pipe could fill the tank in 13.09 hours.

Systems involving second-degree equationsOpen ê Close

23.An engineer is to design a rectangular computer screen with a 19 inch diagonal and a 175 square-inch area. Find the dimensions of the screen to the nearest tenth of an inch.

Input @

MDPlot @80<, 8x, 0, 15 <, PlotRange → 80, 12 <,


Text @"19 inch" , 88, 7 <D, Line @8815, 0 <, 815, 12 <<D,

Line @880, 12 <, 815, 12 <<D, Text @"S = 175 square −inch" ,

85, 4 <D, Text @"x" , 87, 1 <D, Text @"y" , 81, 7 <D < D;



2 4 6 8 10 12 14x

2

4

6

8

10

12y

19 inch

S = 175 square-inch

x

y

Let x = the width y = the height 19 = diagonal area = 175 square-inch areathe formula for the area of a rectangle x * y = 175

Pythagorean theorem x 2 + y2 = 192

Input @

Æ MDRealOnly ;

Clear @x, y D;

NSolve A9x2 + y2 == 192, x y � 175=, 8x, y < E88x → 11.674, y → 14.9906 <, 8x → −11.674, y → −14.9906 <,

8x → 14.9906, y → 11.674 <, 8x → −14.9906, y → −11.674 <<

Note: The dimension can't have negative values.

Input @ x = 14.990604021151215`

14.9906

Input @ y = 11.673979230795688`

11.674

Answer: The dimensions are 11.7 in and 15.0 in.

24. Find two numbers such that their sum is 3 and their product is 1.



Let x = the first number y = the second number sum of numbers is x + y = 3product of numbers is x * y = 1


Clear @x, y D;

Solve @8x + y � 3, x y � 1<, 8x, y < D

::x →1

2J3 − 5 N, y →

1

2J3 + 5 N>,

:x →1

2J3 + 5 N, y →

1

2J3 − 5 N>>


2 J3 + 5 N and Å

1

2 J3 - 5 N.

25. Find two numbers such that their difference is 1 and their product is 1. (Let x be the larger number and y the smaller number.)

Let x = the first number y = the second number difference of numbers is x - y = 1product of numbers is x * y = 1


Clear @x, y D;

Solve @8x − y == 1, x y == 1<, 8x, y < D



::x →1

2J1 − 5 N, y →

1

2J−1 − 5 N>,

:x →1

2J1 + 5 N, y →

1

2J−1 + 5 N>>


2 J1 - 5 N and Å

1

2 J-1 - 5 N or Å

1

2 J1 + 5 N and Å

1

2 J-1 + 5 N.

26. Find the leghts of the legs of a right triangle with an area of 30 square meters if its hypotenuse is 13 metres long.

Input @

MDPlot A80<, 8x, 0, 12 <, PlotRange → 8−0.1, 5.1 <,


Text @"13 m" , 86, 3 <D, Text A"S = 30 m2" , 83, 2 <E,

Text @"x" , 85, 0.4 <D, Text @"y" , 80.6, 2.5 <D, PointSize @0.025 D,

Point @80, 0 <D, Point @812, 0 <D, Point @80, 5 <D =E;

2 4 6 8 10 12x

1

2

3

4

5

y

13 m

S = 30 m2

x

y

Let x = the length of the first leg of a right triangle y = the length of the second leg of a right trianglearea of a right triangle 30 square metershypotenuse of a right triangle 13 meters

the formula for the area of a right triangle Å1

2 x * y = 30

Pythagorean theorem x 2 + y2 = 132



Input @

Æ MDRealOnly ;

Clear @x, y D;

Solve B:x2 + y2 == 132,1

2 x y == 30>, 8x, y <F

88x → −12, y → −5<, 8x → −5, y → −12<, 8x → 5, y → 12<, 8x → 12, y → 5<<

Note: The lenght of the legs of a right triangle can't have negative values.

Answer: The lenghts are 5 meters and 12 meters.

27. Find the dimensions of a rectangle with an area of 32 square meters if its perimeter is 36 meters long.

Input @

MDPlot A80<, 8x, 0, 16 <, PlotRange → 80, 2 <,


Text A"S = 32 m2" , 85, 0.8 <E, Line @8816, 0 <, 816, 2 <<D,

Line @880, 2 <, 816, 2 <<D, Text @"O = 36 m", 88, 1.8 <D,

Text @"x" , 81, 1 <D, Text @"y" , 87, 0.2 <D = E;

2.5 5 7.5 10 12.5 15x

0.25

0.5

0.75

1

1.25

1.5

1.75

2y

S = 32 m2

O = 36 m

x

y

Let x = the first dimension of a rectangle y = the second dimension of a rectanglearea of a rectangle 32 square metersperimeter of a rectangle 36 metersthe formula for the area of a rectangle x * y = 32the formula for the perimeter of a rectangle 2 x * 2 y = 36




Clear @x, y D;

Solve @8x y == 32, 2 x + 2 y == 36<, 8x, y <D88x → 2, y → 16<, 8x → 16, y → 2<<

Answer: The dimensions are 2 m and 16 m.

28.An artist is designing a logo for a business in the shape of a circle with an inscribed rectangle. The diameter of the circle is 6.5 meters, and the area of the rectangle is 15 square meters. Find the dimensions of the rectangle.

Input @

Clear @x, y D;

ImplicitPlot B:Hx − 3L2 + Hy − 1.25 L2 �6.5

2

2

>, 8x, −2, 7 <,

PlotRange → 8−2, 5 <, Epilog → 9Purple, Line @886, 0 <, 86, 2.5 <<D,

Text @"r = 6,5 m" , 83.5, 1.8 <D, Line @880, 2.5 <, 86, 2.5 <<D,

Text @"S" , 83, 1 <D, Line @880, 0 <, 86, 2.5 <<D, PointSize @0.025 D,

Point @83, 1.25 <D, Text A"S = 15 m2" , 81, 1.8 <E,

Text @"x" , 83, 0.3 <D, Text @"y" , 80.2, 1.2 <D, Point @80, 0 <D,

Point @86, 0 <D, Point @86, 2.5 <D, Point @80, 2.5 <D=F;



1 2 3 4 5 6x

-2

-1

1

2

3

4

5y

r = 6,5 m

S

S = 15 m2

x

y

Let x = the first dimension of a rectangle y = the second dimension of a rectangle6,5 meters is diameter of the circle and also of the rectangle 15 square meters is area of the rectanglearea of a rectangle x * y = 15

Pythagorean theorem x 2 + y2 = 6, 52

Input @

Æ MDRealOnly ;

Clear @x, y D;

Solve A9x2 + y2 == 6.5 2, x y == 15=, 8x, y < E88x → −6., y → −2.5 <, 8x → −2.5, y → −6. <,

8x → 2.5, y → 6. <, 8x → 6., y → 2.5 <<

Note: The dimensions of the rectangle can't have negative values.

Answer: The dimensions are 2.5 meters and 6 meters.



FunctionsOpen êClose Print

Linear Functions theoryOpen ê Close

Definition of Linear Function

A function f is a linear function if f(x) = ax + b , for real numbers a and b.The graph of a linear function is a straight line.The domain of the function is all possible "x" values that satisfy the equation. The range is all possible "y" values that satisfy the equation.

For example:

Input @Clear @f, x D;

f @x_D = 2 x + 1;

Input @

Plot @8f @xD<, 8x, −3, 3 <, PlotLegend → 8"f @xD" <,

LegendPosition → 81.1, −0.4 <, PlotRange → 8−3, 3 <,

AspectRatio → 1, ImageSize → 600D



-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3y

The domain and the range of this fuction are all real number. D=R, Y=R

Zeros (Roots)The zero of a function is the X-value for which Y, the value of the function is zero.


f @x_D = 3 x − 4;

Note: We can use FindRoot to calculate the root.

Input @ FindRoot @f @xD, 8x, 0 <D8x → 1.33333 <

Note: We can also solve the equation f[x]=0. Now we can see that the root is x=1.(3).

Input @

MDPlot @8f @xD<, 8x, −2, 2 <,

Epilog → 8Red, PointSize @0.025 D, Point @81.333, 0 <D<,



AspectRatio → 1, MDPlotLegend → True,

PlotRange → 8−2, 2 <, ImageSize → 600D

-2 -1 1 2x

-2

-1

1

2y

Function monotonicityLinear function is increasing function if a is positive.Example:


f @x_D = 2 x + 1;

Input @MDPlot @8f @xD<, 8x, −10, 10 <, MDPlotLegend → True,

AspectRatio → 1, PlotRange → 8−10, 10 <, ImageSize → 600D



-10 -5 5 10x

-10

-5

5

10y

Linear Function is decreasing if a is negative.Example:


f @x_D = −2 x + 1;

Input @MDPlot @8f @xD<, 8x, −10, 10 <, AspectRatio → 1,

PlotRange → 8−10, 10 <, MDPlotLegend → True, ImageSize → 600D



-10 -5 5 10x

-10

-5

5

10y

Linear function is constant function if a=0.Example:


f @x_D = 1;

Input @MDPlot @8f @xD<, 8x, −10, 10 <, AspectRatio → 1,




-10 -5 5

-10

-5

5

10y

'a' and 'b' coefficientsCoefficient 'a' of linear function says how line should be sloped to x-axis.If 'a' is negative then function is decreasing. If 'a' is positive then function is increasing.

Input @

Clear @f, g, x D;

f @x_D = −5 x + 3;

g@x_D = 5 x + 3;



Input @MDPlot @8f @xD, g @xD<, 8x, −5, 5 <,

AspectRatio → 1, MDPlotLegend → True, ImageSize → 600D

-4 -2 2 4x

-20

-10

10

20

y

Coefficient 'b' of linear function says how line should be moved on y-axis.

Input @

Clear @f, g, h, x D;

f @x_D = −5 x + 3;

g@x_D = −5 x + 6;

h@x_D = −5 x − 8;

Input @MDPlot @8f @xD, g @xD, h @xD<, 8x, −6, 6 <, AspectRatio → 1,




-6 -4 -2 2 4 6

-6

-4

-2

2

4

6y

Note: If only 'b' coefficient is changed then lines are parallel.

Linear Functions examplesOpen ê Close

1. The store-house collected 21 tones of potatoes. Every day it gives 150 kilograms of potatoes. Draw a graph of the function defining the connection between the number of kilograms of potatoes remaining in storage and time.



Note: We have to define function.


f @x_D = 21 000 − 150 x;

Note: Now we can plot the function.

Input @ MDPlot @8f @xD<, 8x, 0, 100 <, ImageSize → 500D

20 40 60 80

8000

10000

12000

14000

16000

18000

20000

y

2. From the swimming pool containig 1000m3 of water, 10m3 floats out each minute. Show the volume of remaining water as the function of time t. Plot a graph of this function.



Note: We have to define and plot the function.

Input @Clear @f, t D;

f @t _D = 1000 − 10 t;

Input @MDPlot @8f @t D<, 8t, 0, 100 <,

AxesLabel → 8"t" , "f @t D" <, ImageSize → 500D

20 40 60 80

200

400

600

800

1000

f@tD

3. Metal rod in ambient temperature 0°C is 15m long. B y increasing the temperature by 1°C the rod is 0,15mm longer. Show t he length of the rod as the temprature function. In which ambient temperature the rod is 0.6cm longer?



Note: We have to define the function.

Input @Clear @f, t D;

f @t _D = 0.015 t + 1500;

Note: t is the temperature of the rod, f[t] is the length of rod.

Input @

"Switch Solve, MDRealOnly, Pure Solve" ;

Clear @xD;

MDRealOnly @Solve @81500.6 � f @t D<, 8t <DD88t → 40. <<

Note: We can use Solve to find out the solution or we can plot the function and read the solution.

Input @

MDPlot @8f @t D<, 8t, 10, 100 <,

Epilog → 8Red, PointSize @0.025 D, Point @840, 1500.6 <D<,

AxesLabel → 8 "t" , "f @t D" <, ImageSize → 500D



40 60 80

1500.4

1500.6

1500.8

1501.0

1501.2

1501.4

f@tD

Answer: Metal rod will be 0.6cm longer in temeprature 40°C .

4.

From medical observations appears that time t[h] of human (aged to 18) daily

sleep time is shown in the function t=8+18 - n

2where n is age in years. In which

age man sleeps less than 10 hours per day, on average?



Note: We have to solve this equation.

Input @

Clear @t, n D;

Solve B:10 � 8 +18 − n

2>, 8n<F

88n → 14<<

Answer: Man after the age of 14 years sleeps on average less than 10 hours per day.

5. In certain city, price of bus ticket is function of the number of traveled bus stops. Write the function formula assuming that it is linear function and plot it for bus line in which between ending and starting bus stop there are 18 other.


f @x_D = k ∗ x;

Input @

Clear @fTable, x D; fTable @x_D : = x;

start = 0;

stop = 19;

step = 1;

data = Chop@Table @8x, N @fTable @xDD<, 8x, start, stop, step <DD;

MDShowTable@data, 8"x value" , fTable @xD<D;

x value x0 0

1 1.

2 2.3 3.

4 4.5 5.

6 6.

7 7.8 8.

9 9.

10 10.



11 11.

12 12.13 13.

14 14.15 15.

16 16.

17 17.18 18.

19 19.

Input @MDPlotData @data, FrameLabel → 8"X Values" , "Y Values" <,

PlotStyle → 8Red, PointSize @.02 D<, PlotJoined → False D

0 5 10 150

5

10

15

X Values

YV

alue

s

19 Data Points

Answer: This function is y=kx, for example y=x.

6.

The cost of hiring a store-house A can be counted as function k[x]=250+15x and cost of hiring store-house B can be counted as function h[x]=90+45x, where x is number of weeks of hiring. Make plots of these functions and find out which one is more attractive and for how many weeks.



Note: We have to define both functions.

Input @

Clear @k, x, h D;

k@x_D = 250 + 15 x;

h@x_D = 90 + 45 x;

Note: Now we can plot both functions.

Input @MDPlot @8k@xD, h @xD<, 8x, 0, 10 <, AxesLabel → 8"weeks" , "price" <,

MDPlotLegend → True, ImageSize → 700, PlotRange → 80, 700 <D

0 2 4 6 8 10

100

200

300

400

500

600

700price

Note: We should find the point where functions are crossing.



Clear @xD;

N@Solve @8k@xD � h@xD<, 8x<DD88x → 5.33333 <<

Answer: First store-house is more attractive to hire for longer than 5,3 weeks.

7. Spring's length y[cm] is a function of weighting with mass x[kg] declared by function y=2x+20. Allocate values of mass x, which you can weight spring, so the length would be between 30cm and 70cm.

Note: We have to find out weight when length of the line is 30 and 70 cm long.

Input @Clear @x, y D;

Solve @830 � 2 x + 20<, 8x<D88x → 5<<

Input @ Solve @870 � 2 y + 20<, 8y<D88y → 25<<

Answer: Weight should be between 5 and 25 kg.

8.

It is known that from 90kg of dry linum it is possible to make 8kg linen canvas. How many kg of linum is needed to make 6kg linen canvas? Write a function declaring amount of gained linen canvas depending on the amount of linum taken to production?



Note: We have to create proportion.90kg - 8kgxkg - 6kg

Input @

Clear @xD;

NBSolve B:x �90 × 6

8>, 8x<FF

88x → 67.5 <<

Answer: To make 6kg linen canvas we have to use 67.5 linum.

Note: We have two points A=(67.5, 6) and B=(90, 8).xA=67.5

yA=6

xB=90

yB=8

Input @ xA = 67.5;

Input @ yA = 6;

Input @ xB = 90;

Input @ yB = 8;

Input @Clear @x, y D;

Solve @8 Hy −y AL ∗ HxB −x AL == HyB −y AL ∗ Hx −x AL<, 8 y < D88y → 0.0444444 H135. + 2. H−67.5 + xLL<<

Input @ FullSimplify @y = 0.044444444444444446` H135.` + 2.` H−67.5` + xLLD0. + 0.0888889 x




f @x_D = 0.08888888888888889 x;

Input @MDPlot @8f @xD<, 8x, 0, 100 <,

AxesLabel → 8 "x" , "f @xD" <, ImageSize → 500D

20 40 60 80

2

4

6

8

f@xD

Answer: This function is f[x]=0.0(8)x.

Quadratic functions - theoryOpen ê Close

Definition of a Quadratic FunctionQuadratic functions are any functions that may be written in the form

y = ax2 + bx + c where a, b, and c are real coefficients and a ∫ 0.

The expression ax2 + bx + c is called 2nd degree polynomial.

Graph of a Quadratic FunctionThe graph of a quadratic function is called a parabola .



Examples:


f @x_D = x2;



-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3y


f @x_D = −x2;





-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3y

Maximum, Minimum, Vertex of a Quadratic Function

The Vertex is the highest (maximum) or lowest (minimum) point of the graph.The vertex is the highest point on a parabola if the parabola opens down (a<0), it is the lowest point if the parabola opens up (a>0).

Input @

Clear @f, g, h, x D;

f @x_D = x2;

g@x_D = 2 x 2 + 3 x − 1;

h@x_D = −0.5 x 2 − x + 3;

Input @MDPlot @8f @xD, g @xD, h @xD<, 8x, −5, 5 <, AspectRatio → 1,




-4 -2 2 4

-5

5

10

y

The x-coordinate of the vertex is x=-b

2 a. The y-coordinate of the vertex is y=-

D

4 a. In

Mathematica we can use FindMinimum or FindMaximum depending on what we are looking for.

Note: D=b2-4ac

Input @ FindMaximum @h@xD, 8x<D83.5, 8x → −1. <<

Note: Coordinates of vertex of function h[x] are (-1,3.5).



Input @ FindMinimum @g@xD, 8x<D8−2.125, 8x → −0.75 <<

Note: Coordinates of vertex of function g[x] are (-0.75,-2.125).

Input @ FindMinimum @f @xD, 8x<D80., 8x → 0. <<

Note: Coordinates of vertex of function f[x] are (0,0).

Input @

MDPlot @8f @xD, g @xD, h @xD<, 8x, −5, 5 <,

PlotLegend → 8"f @xD" , "g @xD" , "h @x" <, PlotRange → 8−4, 4 <,

Epilog → 8Red, PointSize @0.025 D, Point @8−1, 3.5 <D,

Point @8−0.75, −2.125 <D, Point @80, 0 <D<,

MDPlotLegend → True, ImageSize → 800D

-4 -2 2 4

-4

-2

2

4y

Zeros

Function f[x] = ax2 + bx + c (a∫0):

1. For D>0 has 2 different roots: x1=-b - D

2 a and x2=

-b + D

2 a



2. For D=0 has 1 (double) root x=-b

2 a3. For D<0 has no roots

Examples:

Find roots of function f[x]=-2 x2-x+6.We have to calculate D first.

Input @

Clear @a, b, c, ∆, x 1, x 2D;

a = −2;

b = −1;

c = 6;

Input @ Solve A9∆ � b2 − 4 a c =, 8∆<E88∆ → 49<<

Input @ ∆ = 49;

Note: Now we can calculate roots.

Input @ Solve B:x1 �−b − ∆

2 a, x 2 �

−b + ∆

2 a>, 8x1, x 2<F

::x1 →3

2, x 2 → −2>>

Answer: Roots of function f[x] are x1 = Å3

2, x2 = -2.

Forms of a quadratic function

1. f[x] = ax2 + bx + c is called the standard form .

2. f[x] = a(x-x1)(x-x2) is called the factored form , where x1and x2 are the roots of the

quadratic equation.

3. f[x] = aHx - pL2+q is called the vertex form , where p and q are the x and y coordinates of

the vertex.

Transforming vertex form to standard form:

Example: Write y=3Hx - 2L2+4 in standard form.We just have to expand this equation.

Input @ Expand A3 Hx − 2L2 + 4E

16 − 12 x + 3 x 2



Answer: Standard form is y=3 x2-12x+16.

Transforming standard form to vertex form:

Example: Write y=x2-6x+4 in vertex form.We already know a=1. We just need to calculate p and q.

p=-b

2 a and q=-

D

4 a (D=b2-4ac)

Input @

Clear @a, b, c, p, q, ∆D;

a = 1;

b = −6;

c = 4;

Input @ Solve B:p � −b

2 a>, 8p<F

88p → 3<<

Input @ p = 3;

Note: p = 3. Now we need to calculate D.


Input @ ∆ = 20;

Note: D = 20 so now we can calculate q.

Input @ Solve B:q � −∆

4 a>, 8q<F

88q → −5<<

Note: Now we know all needed variables to complete vertex form.

f[x]=aHx - pL2+q

f[x]=Hx - 3L2-5

Transforming factored form to standard form:

To do this we just have to use a "FullSimplify" on the equation, for example:

Write f[x]=- Å1

3(x-3)(x+3) in standard form:



Input @ FullSimplify B−1

3Hx − 3L Hx + 3LF

3 −x2

3

Answer: Standard form of this function is f[x]=- Å1

3x2+3

Transforming standard form to factored form:We have to calculate roots and put them into formula (we already know a).Example:

Write f[x]=5x2 - 15 x in factored form.

Input @

Clear @a, b, c, ∆, x 1, x 2D;

a = 5;

b = −15;

c = 0;


Input @ ∆ = 225;

Note: Now we can calculate roots.


2 a, x 2 �

−b + ∆

2 a>, 8x1, x 2<F

88x1 → 0, x 2 → 3<<

Note: 5(x-0)(x-3)=5x(x-3)

Answer: Factored form of this function is f[x]=5x(x-3).

Quadratic functions - examplesOpen ê Close

9. The number of admissions to hospital on an n-day of the epidemy can be presented in a formula A(n) = 10 +30n-3n2. Find out how many sick people were admissioned to hospital during one day.



Note: First we will find maximium arrived patients in one day - it's a vertex of A(n) function (so also maximum of that function)

Input @Clear @f, n D;

f @n_D = 10 + 30 n − 3 n2;

Input @ FindMaximum @f @nD, 8n<D885., 8n → 5. <<

Answer: The maximum amount of patients in one day was 85

10.On the 9-th March, 2010 a person celebrating their jubilee said:"When I multiply my age of 17 years earlier and my age in 20 years time, I'll get this year, 2010". Which birthday is he/she celebrating?

Note: We must find a pattern first. x is a current age of a birthday

Input @ FullSimplify @Hx − 17L ∗ Hx + 20L � 2010Dx H3 + xL � 2350

Note: The pattern is x(3+x) = 2350. You can write it in form: f[x]=x2+3x-2350. As you



can see it is a quadratic function.

Input @


Clear @xD;

MDRealOnly ASolve A9x2 + 3 x − 2350 � 0=, 8x<EE88x → −50<, 8x → 47<<

Note: As you see, we have 2 answers, but the first one (-50) is incorrect (we can't have a minus age)

Answer: The birthday have 47 years

11.There was a shop and its number of clients can be expressed by a formula d(n) = -2n2+32n-8, where n is the number of days, nœN+ and 1§n§15. Which day was the number of clients the biggest?

Note: First thing to do is define a function

Input @Clear @d, n D;

d@n_D = −2 n2 + 32 n − 8;

Note: Secondly we will find a maximum

Input @ FindMaximum @d@nD, 8n<D8120., 8n → 8. <<

Answer: Most customers came in 8th day.

A bullet moving upwards falls down verticaly after reaching the maximum point of its route.Quadratic function can be used to describe it. For example the

2



function h(t) = -5t2+50t+5 assigns to time t[s] the height h[m] where the bullet is.

a) Plot the graph of that function.b) What is the maximum height which bullet reached?c) How long the bullet was in air?d) After what time the bullet was in the same height which it was launched?

Note: For plot the data graph we will use 'def f' and 'Plot' buttons

Input @Clear @h, t D;

h@t _D = −5 t 2 + 50 t + 5;

Input @MDPlot @8h@t D<, 8t, 0, 11 <,

PlotRange → 80, 140 <, AxesLabel → 8 "t" , "h @t D" <D

0 2 4 6 8 10t

20

40

60

80

100

120

140h@tD

Note: For find the maximum height we will use a 'FindMaximum' function built-in M@th Desktop

Input @ FindMaximum @h@t D, 8t <D8130., 8t → 5. <<

Answer: The maximum height ehich bullet reached is 130 meters



Note: for check how long the bullet was in air we will use a 'Solve' button

Input @


Clear @xD;

N@MDRealOnly @Solve @8h@t D � 0<, 8t <DDD88t → −0.0990195 <, 8t → 10.099 <<

Note: We have two answers - of course the first one (-0.0990195) is incorrect.

Answer: The bullet was in air 10.099 seconds

Note: To check from which height the bullet was launched we must check what value the function have for 0 second

Input @ h@0D5

Note: The bullet was launched from 5 meters. We will use now again a 'Solve' button to check after what time the bullet reach the same height

Input @


Clear @xD;

MDRealOnly @Solve @8h@t D � 5<, 8t <DD88t → 0<, 8t → 10<<

Note: First answer we already have - it's the time when the bullet was launched.

Answer: Bullet reached the same height after 10 seconds

13.

An electric train sets from station A, and stops at station B. It speed vBkm

hF, after

t minutes from leaving the station A is described by the formula: v= Å1

5 t H4 - tL.

Find out the time to cover the distance from station A to station B.



Note: We will define a pattern

Input @

Clear @v, t, x D;

v@t _D =1

5t H4 − t L;

Note: To calculate the time of travel we will use a 'Solve' button.

Input @"Switch Solve, MDRealOnly, Pure Solve" ;

MDRealOnly @Solve @8v@t D � 0<, 8t <DD88t → 0<, 8t → 4<<

Answer: The travel time from station A to B equals 4 minutes

14.

You are the cinema owner and you know that when the ticket price is 10 €, the number of people coming to the cinema is about 100. When the price is increased by one Euro, the number of clients is less by 5 people. Which price is the best for your income?

Note: First we must find a pattern. The price of ticket is 10+x and the number of people is 100 - 5x.


f @x_D = Expand @H10 + xL H100 − 5 xLD



1000 + 50 x − 5 x 2

Note: For a better look we can plot that function

Input @ MDPlot @8f @xD<, 8x, 0, 10 <, AxesLabel → 8 "x" , "f @xD" <D

2 4 6 8 10

1020

1040

1060

1080

1100

1120

f@xD

Note: Now we must find a maximum value of that function

Input @ FindMaximum @f @xD, 8x<D81125., 8x → 5. <<

Answer: For the best income you should set the price of ticket to 5 €.

15.A function defined by a formula f(x) =

-x2 + 6 x + 21

2 describes a worker's

productivity depending on working time x during eight hour working day. This worker starts working at 7 a.m. When is the highest productivity?



Note: We will define a pattern

Input @

Clear @f, x D;

f @x_D =1

2I−x2 + 6 x + 21M

1

2I21 + 6 x − x2M

Input @1

2I21 + 6 x − x2M

Note: Now we will find a maximum

Input @ FindMaximum @f @xD, 8x<D815., 8x → 3. <<

Note: The third hour of work is the most performance hour. So if worker starts his job at 7 a.m. then he reach the best performance in 10 a.m..

Answer: The performance of worker is the biggest at 10 a.m.

16.66 games were played in a chess tournament, in which each player played one game with each competitor. How many people took part in this competition?



Note: First we must find a pattern. Let the x describe a number of all players.

Input @ FullSimplify @x Hx − 1L ê 2 � 66Dx2 � 132 + x

Note: As you can see it's a quadratic function: f(x) = x2-x-132. Now we must find a number of players, so we must

Input @

Clear @x1, x 2D;

a = 1;

b = −1;

c = −132;

Input @ ∆ = b2 − 4 ∗ a ∗ c

529


2 a, x 2 �

−b + ∆

2 a>, 8x1, x 2<F

88x1 → −11, x 2 → 12<<

Note: We have two answers but -11 is incorrect (there can't be -11 number of players on a tournament)

17.A life guard has got a rope and wants to make a special rectangular zone for children with this rope of the biggest possible area. Give the dimensions for a rope 100 meter long.



Note: First we must find a pattern. Let the a and b will the sides of the rectangle

Input @ Clear @a, b, c, f, x DInput @ 2 a + 2 b � 100;

Note: Then b=50-a

Input @ f @a_D = a H50 − aLH50 − aL a

Note: Now we must find a maximum of that function

Input @ FindMaximum @f @cD, 8c<D8625., 8c → 25. <<

Input @"Switch Solve, MDRealOnly, Pure Solve" ;

MDRealOnly @Solve @82 × 25 + 2 b � 100<, 8b<DD88b → 25<<

Answer: The swimming pool should have a dimensions 25x25 for the biggest area

18. Which number added to its square gives the smallest total amount?



Note: Let the x will be te searched number.


f @x_D = x + x2

x + x2

Note: We will use a find minimum function to solve that problem

Input @ FindMinimum @f @xD, 8x<D8−0.25, 8x → −0.5 <<

Answer: This number is -0.5

Physics problemsOpen ê Close

Note: This problems cannot be solved by student of technical school without using Mathematica program

Relativistic mass

19.

In relativistic physics mass of object isn't constant. It is variable from velocity of this mass. Plot the funcion describing mass relation from velocity for starting mass(m0).Conclude the result.m0=500[g]

c=3 * 10 8[m/s]

Note: Getting started we need to declare starting mass(m 0) and also for input

simplification light's speed(c)

Input @ clear @c, v, m 0D; m0 = 500; c = 3 ∗ 108;

Note: Inputing the funcion (f(v))

Input @f @v_D =

m0

1 − NBI �vcM2F



500

1 − 1.11111 × 10−17 v2

Note: As we see program itself put our declared variables to funcion. Now the last thing left to do is ploting the funcion in range from 0 to c.

Input @MDPlot @8f @vD<, 8v, 0, c <,

PlotStyle → 88Red, Thickness @.005 D<<, PlotRange → 80, 1400 <D

0 5.0 µ 107 1.0 µ 108 1.5 µ 108 2.0 µ 108 2.5 µ 108 3.0 µ 108x

200

400

600

800

1000

1200

1400y

Answer: At the funcion graph we see mass relation to velocity. When mass velocity is close to light speed relativistic mass is multiple times higher then stoic mass of object.

20.Find the object velocity at which relativistic mass ratio to stoic mass equals n= 4/3.c=3*10^8[m/s]

Note: First we need to declare our variables: mass ratio,lightspeed

Input @ clear @m0, m, v, c, n D; n = 4 ê 3; c = 3 ∗ 108;

Note: Second we need to input our formula.

Input @ v = f @n_D = NBc ∗ 1 −1

n2F

1.98431 × 108



Answer: Velocity at which mass ration equal 4/3 is 1.98431*108[m/s] and is lower than lightspeed which was easy to guess.

Waves

21.

Equation describing source vibration is y=0.04*sin(600pi*t). Vibrations are in elastic envirioment. Find period of this vibration, kinematic wave equation and describe deviation from balance point which is in distance of 75cm from source after the time of 0,01 s from starting moment.Wave speed v=300m/s.Plot the first equation

Note: First we need to find period from pulsation formula w=2 p

T.

After formula transformation we can find the period. T=2 p

w

Input @

Clear @T, ωD;

ω = 600 π;

T = f @ωD =2 π

ω;

Note: We have found the funcion period and it equals 1/300[s]. Kinematic wave formula

is described by y=Asin(2p(t

T- Åx

l)). We can also evaluate lenght of the wave from

velocity formula v=l

T.

Input @

A = 0.04 ;

t = 0.01;

v = 300;

x = 0.75;

λ = f @v_D = T ∗ v;

Note: Now we have all the data so we can evaluate wave deviation.

Input @ Evaluate BSin B2 π ∗t

T−

x

λFF

1.

Input @ y = A ∗ Sin B2 π ∗t

T−

x

λF

0.04



Note: Last thing to do is to plot the equation y=0.04sin(600pt)

Input @Clear @t D;

f @t _D = 0.04 ∗ Sin @600 π ∗ t D;

Input @ Plot @8f @t D<, 8t, 0, 0.05 <, AxesLabel → 8 "t" , "f @t D" <D

0.01 0.02 0.03 0.04 0.05t

-0.04

-0.02

0.02

0.04f@tD

Answer: Wave deviation equals 0.04 which is amplitude of this wave.

TrigonometryOpen êClose Print

Solved problemsOpen ê Close

1. A boy 1,5 m tall is standing 15 m away from an inflatable clown. He is staring upat a 45° angle. How tall is the clown?



Input @

angle = 45;

oppositeLeg = 15;

adjacentLeg = oppositeLeg ê Tan@angle ê 180 ∗ πD êê N

15.

We calculate distance BC by tan.

Input @15 + 1.5

16.5

The height of the clown is sum distance of BC and the height of the boy.

Answer: The clown is 16,5 m tall.

2. A little kid is flying a balloon. The string of the balloon makes an angle of 30degrees with the ground. If the height of the balloon is 12 m, find the length ofthe string that the boy has used.



Input @

Æ c ê Sin@gD ;

a = 12;

α = 30;

β = 90;

b =a

Sin A α∗π180

E∗ Sin Bβ ∗ π

180F êê N

24.

To find the lenght of the string, we use the law of sines, because we know two angles and the height of the balloon.

Answer: The lenght of the string is 24 m.

3. Two ships, A and B leave Mersey Docks (C) at the same time. Ship A travels ata bearing of 120° and ship B travels at a bearing o f 100°. After 1 hour ship A hastravelled 25 km and angle CBA is 130°. Find the spe ed of ship B.



120°-100° = 20° �> the angle ACB is 20°180° - 130° -20° = 30° �> the angle CAB is 30°

We use the law of sines.

Input @

Æ c ê Sin@gD ;

b = 25;

β = 130;

α = 30;

a =b

Sin A β∗π

180E

∗ Sin Bα ∗ π

180F êê N

16.3176



Answer: The speed of ship B was 16,32 km/h .

4. The sides of triangle ABC are 39 cm, 42 cm, 45 cm. The second longest heightof triangle is 36 cm. What is the shortest height?

We know: a = 42 cm, b = 39 cm, c = 45 cm, h = 36 cm

In order to find the angle b, we use the law of sines.

Input @

Æ Sin@gD êc ;

δ = 90;

c = 45;

h = 36;

β = ArcSin Bh ∗Sin A δ∗π

180E

cF ∗

180

πêê N

53.1301

Now we use the law of sines to measure x.

Input @

Æ a ê Sin@aD ;

a = 42;

γ = 90;

β = 53.1301 ;

x =a

Sin A γ∗π

180E

∗ Sin B β ∗ π

180F êê N



33.6

Answer: The shortest height of triangle is 33,6 cm .

5. Two towers face each other separated by a distance of 30 m. As seen from thetop of the first tower, the angle of depression of the second tower´s base is 60°and that of the top is 30°. What is the height of t he second tower?

First, we find x and y by sine. The height of the second tower is x+y.

Input @

angle = 30;

adjacentLeg = 30;

oppositeLeg = adjacentLeg ∗ Tan@angle ê 180 ∗ πD êê N

17.3205

Input @ x = 17.32050807568877

17.3205

Input @

angle = 60;

adjacentLeg = 30;




51.9615

Input @ y = 51.96152422706631

51.9615

Input @ h = x + y

69.282

Answer: The height of the second tower is about 69 m.

6. A plane is flying above a place A. Danny can see the plane from a place B (B is2 400 m away from A) at an angle of elevation of 54°20'. What is the height ofthe plane above the Earth?

Angle ABC is 54°20'. We can calculate AC by tan.

Input @

angle = 54.33;

adjacentLeg = 2400;


3343.65



Answer: The plane is flying 3343,65 m above ground.

7. The walker is standing 150 m from Orava castle. He can see the top of thecastle at 37° What is the height of the castle?

Input @

Æ a ê Sin@aD ;

AB = 150 ;

γ = 53;

β = 37;

b =AB

Sin A γ∗π

180E

∗ Sin B β ∗ π

180F êê N

113.033

We know AB and all angles, so we use the law of sines to calculate the height of the castle.



Answer: The height of the castle is 113,033 m.

8.

The first person is standing on the bank of the Orava dam and is 900 m far fromthe second person, who is standing on the other side of the dam. The thirdperson can see the first one from a glider at 70° a nd the second at 60°.Calculate the height of the glider.

Input @ 20 + 30

50

This is the third angle in triangle 12A

Input @

Æ a ê Sin@aD ;

c = 900 ;

γ = 50;

β = 60;

b =c

Sin A γ∗π

180E

∗ Sin B β ∗ π

180F êê N

1017.46

We need to know distance 1A. We can use the law of sines.



Æ c ê Sin@gD ;

b = 1017.4642873624224` ;

γ = 90;

α = 70;

a =b

Sin A γ∗π

180E

∗ Sin B α ∗ π

180F êê N

956.104

We use the law of sines to calculate the height of the glider.

Answer: The height of the glider is 956,104 m.

9.

The statue in Klin ,,Rio de Klin" is 9,5 m tall. Two people are standing on theright and left side of the statue. The person who is standing on the right can seethe top of the statue at 65°. The distance of perso n on the left from the statue is1,5 times smaller than the distance of person on the right. At what angle can theperson on the left see the top of the statue?

Æ a ê Sin@aD ;

v = 9.5 ;

β = 65;



65;

δ = 25;

SB =v

Sin A β∗π

180E

∗ Sin B δ ∗ π

180F êê N

4.42992

We calculate distance of person on the right side from the statue by sin.

Input @ 4.42992 ê 1.5

2.95328

The distance of person on the left is 1.5 times smaller, so it is 2,95328 m.

We use cosine function to find angle a.

Input @

oppositeLeg = 9.5;

adjacentLeg = 2.95328;

angle = ArcTan @oppositeLeg ê adjacentLeg D ∗ 180 ê π êê N

72.731

Answer: The person on the left can see the top of the statue at 72,731°.

10.

On shore of the river there is a building. From one window of this building youcan see the point on the other side of the river under 25,3°. From other windowwhich is about 12 metres higher you can see this point under 37,5°. How wide isthat river?



At first we have to calculate the size of angle AW2P.

Input @

angle1 = 37.5 ;

angle2 = 90;

AW2P= 180 − angle1 − angle2

52.5

Now we can find out W1P by the law of sines, angle W1PW2 = 37.5°-25.3°=12.2°

Input @

Æ a ê Sin@aD ;

b = 12;

β = 12.2 ;

γ = 52.5` ;

c =b

Sin A β∗π

180E

∗ Sin Bγ ∗ π

180F êê N

45.0503

W1P = 45.0503 m. Lets calculate the size of angle AW1P and after that we can use the law of sines again to find out the width of the river.



Input @

angle1 = 25.3 ;

angle2 = 90;

AW1P= 180 − angle1 − angle2

64.7

Input @

Æ b ê Sin@bD ;

W1P= 45.05027447408226` ;

α = 90;

AW1P= 64.69999999999999` ;

c =W1P

Sin A α∗π180

E∗ Sin BAW1P∗ π

180F êê N

40.7292

Answer: The width of the river is 40,7292 m.

11.

The area lighted by a circular lamp has a conic shape. The angle of top of thiscone is 36°. The lamp is hanging on a mast which is 25 m high. An anglebetween an axis of the mast and an axis of the top angle is 42°. How long isthe lighted stretch of the road?

At first we can calculate the distance between A and C by tangens function in right triangle ADC. Angle ADC is 42° + 18° = 60°



angle = 60;

adjacentLeg = 25;


43.3013

AC = 43,3013 m. Now we can find out AB by tangens function, too. Angle ADB is 42° - 18° = 24°.

Input @

angle = 24;

adjacentLeg = 25;


11.1307

AB = 11,1307 m.

Input @BD = 43.30127018922193` − 11.130717132713402`

32.1706

Answer: The length of the lighted strech is 32,1706 m.

12.Target C is watched from two gunnery positions A and B which are 296 m fromeach other. We know that angle BAC is 52 ° and ABC is 44 °. How far is A fromC?



g=180°- (52°+ 44°)

Input @

angle1 = 52;

angle2 = 44;

γ = 180 − angle1 − angle2

84

Input @

Æ c ê Sin@gD ;

c = 296 ;

γ = 84;

β = 44;

b =c

Sin A γ∗π

180E

∗ Sin B β ∗ π

180F êê N

206.751

Answer: The distance between A and C is 206,751 m.

13. In the triangle ABC the ratio of its sides is a:b:c = 3 :1:2 and the angle ABC is60°. Solve: sin(ABC) + sin (BCA) = ?

Using Pythagoras´ Theorem we can show that ABC is a right triangle.



Input @

firstLeg = 1;

secondLeg = 3 ;

hypotenuse = firstLeg 2 + secondLeg 2 êê N

2.

Input @

hypotenuse = 2;

adjacentLeg = 3 ;

α = ArcCos @adjacentLeg ê hypotenuse D ∗ 180 ê π êê N

30.

Input @β = 60

γ = 90

60

90

Input @ Sin B β ∗ π

180F + Sin B γ ∗ π

180F

Input @ 1 +3

2êê N

1.86603

Answer: sin(ABC) + sin (BCA) is 1,86603.

14.

The army convoy is marching along a straight road. We are in the place A. The distance between the begining of the army convoy (B) and A is 14 350 m. From the end of the convoy (C) to A 13 840 m. The angle BAC is 51°. How long is the army convoy?



We know: b=13840 c=14350 a=51°

We calculate the length of the army convoy by the law of cosines.

Input @

b = 13 840;

c = 14 350;

α = 51;

a = b2 + c2 − 2 ∗ b ∗ c ∗ CosBα ∗π

180F êê N

12 144.8

The army convoy is long 12 144,8 m.

15.

Two ships are floating from the island ,,Slanica". When the first ship crosses 30metres, it can see the island ,,Slanica" at angle of elevation of 38°. After 20metres, the second ship can see the island ,,Slanica" at angle of elevation of46°. Calculate the distance of two ships from each other.



Input @

angle1 = 38;

angle2 = 46;

γ = 180 − angle1 − angle2

96

We know two angles - 38° and 46°, so we can easily find the third angle - 180° - ( 38° + 46° ) = 96°

Input @

a = 20;

b = 30;

γ = 96;

c = a2 + b2 − 2 ∗ a ∗ b ∗ CosBγ ∗π

180F êê N

37.7549

We calculate the distance of two ships by the law of cosines, because we know two sides and an angle.

Answer: The distance of two ships is 37,76 metres .



We can see the top of tower from a place A at an angle of elevation of 37°.When we get 50 metres nearer to the bottom of the tower - B, the angle ofelevation will be 58°. How high is the tower?

Input @β = 58;

δ = 180 − β

122

Input @

α = 37;

δ = 122;

γ = 180 − Hα + δL21

We know all angles in the triangle ABV.

Distance AV we can calculate by the law of sines



Input @

Æ a ê Sin@aD ;

v = 50;

γ = 21;

δ = 122 ;

AV =v

Sin A γ∗π

180E

∗ Sin B δ ∗ π

180F êê N

118.321

We calculate the height of the tower by sine function

Input @

angle = 37;

hypotenuse = 118.321;

oppositeLeg = Sin @angle ê 180 ∗ πD ∗ hypotenuse êê N

71.2074

Answer: The tower is 71,2074 metres high.

17.Two trains are coming out of a station following two straight lines which makethe angle of 156°. The speed of the first train is 13 m/s, the speed of the secondone is 14,5 m/s. How far from each other will these two trains be in 5,5 minutes?



Input @ 5.5 ∗ 60

330.

Now we claculate how many seconds there are in 5,5 minutes.

Input @ 13 ∗ 330

4290

Input @ 14.5 ∗ 330

4785.

We calculate distance 1S and 2S by formula: s(distance)=v(speed)·t(time).

Input @

b = 4785;

c = 4290;

α = 156;

a = b2 + c2 − 2 ∗ b ∗ c ∗ CosBα ∗π

180F êê N

8877.29

Distance 12 we can calculate by the law of cosines.



Answer: After 5,5 minutes two trains will be 8877,29 metres apart.

18.The plane is flying 3000 m above the ground. At first, a person standing on the ground can see this plane at an angle of 25° , afte r a while, at an angle of 50°. How far did the plane fly during that time?

At first we have to calculate AL1 by cosine function in the right triangle ABL1. Angle AL1B is 90° - 25° = 75°

Input @

angle = 75;

adjacentLeg = 3000;

hypotenuse = adjacentLeg êCos@angle ê 180 ∗ πD êê N

11 591.1

AL1 = 11591.1 m. We can use the law of sines to find out L2L1. Angle AL2L1 = 180° - 50° = 130° . And we have to calculate the size of a ngle L2AL1.

Input @

angle1 = 130;

angle2 = 25;

L2AL1 = 180 − angle1 − angle2

25



Æ b ê Sin@bD ;

c = 11591.109915468822` ;

γ = 130 ;

α = 25;

x =c

Sin A γ∗π

180E

∗ Sin B α ∗ π

180F êê N

6394.69

Answer: The plane flew over 6394,69 m during that time.

19. From two places which are 3100 m far from each other two people can see a cloud in the sky at angles of 65° and 79° . How hig h is the cloud?

In the picture you can see that angle MNA is 79° , too and angle NMA is 65° . Now lets calculate the third angle in the triangle MNA.

Input @

angle1 = 79;

angle2 = 65;

MAN= 180 − angle1 − angle2

36

Now we can find out NA by the law of sines.



Æ b ê Sin@bD ;

c = 3100 ;

γ = 36;

α = 65;

a =c

Sin A γ∗π

180E

∗ Sin B α ∗ π

180F êê N

4779.9

NA = 4779.9 m. Lets find out how high the cloud is. We can use sine function in the right triangle BNA.

Input @

angle = 79;

hypotenuse = 4779.899000282544 ;̀

adjacentLeg = Sin @angle ê 180 ∗ πD ∗ hypotenuse êê N

4692.08

Answer: The cloud is 4692,08 m above the ground.

20.Tourists were following a map, from place A to B they went 2 km straight north,from B to C at a bearing of 53° 13 km and from C to D at a bearing of 90° 4 km.What is the direct distance from A to D?



We know a=53°, b=90°, −BEC=d=90°. So −BCE=g must be 37°.

Input @

Æ c ê Sin@gD ;

BC= 13;

δ = 90;

γ = 37;

BE =BC

Sin A δ∗π180

E∗ Sin B γ ∗ π

180F êê N

7.8236

Input @

Æ c ê Sin@gD ;

BC= 13;

α = 53;

δ = 90;

EC=BC

Sin A δ∗π180

E∗ Sin B α ∗ π

180F êê N

10.3823



We used the law of sines to find BE, EC. In order to find AD, we use the right triangle ADE in which AE=AB+BE=9,236 and ED=EC+CD=14,3823.

Input @

firstLeg = 9.236;

secondLeg = 14.3823;

hypotenuse = firstLeg 2 + secondLeg 2 êê N

17.0925

Answer: The distance from A to D is 17,0925 km .

21.

Lomnický š�tít is 2632 m a. s. l.. There is a cable railroad from Skalnaté pleso(L) to Lomnický štít (V), see the picture. What is the altitude of Skalnaté pleso?How long does it take to get from L to V, if the speed of the cable railroad is 4,1m/s?



We can find LV by the law of sines.

Input @

Æ b ê Sin@bD ;

XL = 1653 ;

γ = 70;

α = 90;

LV =XL

Sin A γ∗π

180E

∗ Sin B α ∗ π

180F êê N

1759.09

Input @

hypotenuse = 1759.0858579026826 ;̀

givenLeg = 1653;

leg = hypotenuse 2 − givenLeg 2 êê N

601.643

Skalnate pleso is 2632 - 601.643= 2030,357 m a.s.l.

t (time) = s (distance) / v (speed)

Input @

LV = 1759.09;

v = 4.1;

t = LV ê v

429.046

Input @ 429.04634146341465` ê 60

7.15077

Answer: The time is about 7,15 minutes and Skalnate pleso is 2030,357 m a.s.l .



ProbabilityOpen êClose Print

Calculation TheoremsOpen ê Close

Definitions

Sample SpacesA sample space is the set of all possible outcomes.

Example 1. Consider the experiment of flipping two coins. It is possible to get 0 heads, 1 head, or 2 heads. Thus, the sample space could be the set {0, 1, 2}. Another way to look at it is flip { HH, HT, TH, TT }.In the table the 1st flip is in the horizontal and the 2nd flip in the vertical line and the four outcomes are in black.

Ñ H T

H HH HT

T HT TT

The second way is better because each event is as equally likely to occur as any other.

A classical probabilityThe probability of an event occurring is the number of cases favorable for the event n (E), over the number of total outcomes possible n (S). This is only true when the events are equally likely.

P(E) = n HELn HSL

Empirical ProbabilityEmpirical probability is based on observation. The empirical probability of an event is the relative frequency of a frequency distribution based upon observation.

P(E) = Åf

n

Example 2. This time you are rolling two dice. The sums are { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }. However, each of these aren't equally likely. The only way you can get a sum 2 is to roll a 1 on both dice, but you can get a sum of 4 by rolling a 1-3, 2-2, or 3-1. The following table illustrates a better sample space for the sum obtain when rolling two dice. In the lowest horizontal row are the points of the first dice and in the most left column are the



points of the second dice. The sum of the two dice are counted in the cells.

6 7 8 9 10 11 12

5 6 7 8 9 10 11

4 5 6 7 8 9 10

3 4 5 6 7 8 9

2 3 4 5 6 7 8

1 2 3 4 5 6 7

results 1 2 3 4 5 6

Now it's easy to find for example the propability to get sum 4.

P(" the sum is 4") = 3

36=

1

12.

In the next table we have a probability distribution:

Sum Relative frequence

2 1

36

3 2

36

4 3

36

5 4

36

6 5

36

7 6

36

8 5

36

9 4

36

10 3

36

11 2

36

12 1

36

Total 36

36= 1

Example 3. The propability to get sum between 4 and 8 is 3

36+

4

36+

5

36+

6

36+

5

36=

23

36.

In the next table you find the running total of the relative frequencies or the cumulative



frequency divided by the total frequency.Sum Cumulative Relative Frequency

2 1

36

3 3

36

4 6

36

5 10

36

6 15

36

7 21

36

8 26

36

9 30

36

10 33

36

11 35

36

12 36

36

Again, find the propability to get sum between 4 and 8.

1. Complementation theorem

For the complementary event A of event A exists a theorem

P(A) = P(A doesn't happen) = 1-P(A)



Example 4.The odds for a serial producted single-use item being malfunctional is 1.5% What is the probability for having at least one malfunctioning product in a set of hundred?

Because P(product is malfunctional) = 1.5%, isP(product is functional) = 98.5%

Event "at least one is malfunctional" considers the possibility that there are 1 or 2 or 3 or ... or 100 malfunctioning products. The complementary event is "there are no malfunctional products" ie. "all products are functional".

P(at least one is malfunctional) = 1 - P(all products are functional) = 1 -

0.985100 = 0.7793... º 78%

2. Multiplication theorem

If events A and B do not affect each other in any way, being independant, there is a possibility that the events A and B take place

P(A and B) = P(A)×P(B)

If the events A and B are dependant of each other the multiplication theorem is valid in the form

P(first A, then B) = P(A)×P(B, when A has already taken place)

Example 5.A die is thrown three times. The results are independant of each other. The odds that the result is a series of 1, 2 and 3 are thus

P(1,2,3) = Å1

6× Å

1

6× Å

1

6 =

1

216 º 0.5%

Three tags are taken at random from a box in which contains the number tags 1, 2, 3, 4, 5 and 6. The results are not independant of each other. The probability for the result of three consequent picks being the series 1, 2 and 3 is

P(1,2,3) = Å1

6× Å

1

5× Å

1

4 =

1

120 º 0.8%



1. The pincode of a bank card is four-digit. What are the odds that a person who only remembers the last digit will succeed at entering the code on his first try?

Input @ P HAL =1

10×

1

10×

1

10× 1

1

1000

2.

Four cards are taken at random from a deck of cards. Count the probability fora) All four being acesb) Not a single one being an acec) Atleast one being an ace

a) A = "All are aces"

Input @

P HAL =4

52×

3

51×

2

50×

1

49

=1

270 725êê N

3.69379 × 10−6

b) B = "Not a single one is an ace"

Input @

P HBL =48

52×

47

51×

46

50×

45

4938 916

54 145êê N

0.718737

c) B = "Atleast one is an ace"

P(B) = 1 - P(B)

Input @

P HB¯L = 1 −

38 916

54 145

15 229

54 145êê N

0.281263

Answer:

a) 1

270 725º 3.69379× 10−6º 0,0000037

b) 38 916

54 145º 0.718737º 0,719



c) 15 229

54 145º 0.281263 º 0,281

3. Two balls are taken at random from a box which contains 3 red, 4 blue and 5 black balls. What is the probability for them being the same color?

P("both are red") + P("both are blue") + P("both are black) =

Input @3

12

2

11+

4

12

3

11+

5

12

4

11

Input @19

66êê N

0.287879

Answer: 19

66º 0.287879 º 0,288

3. Tree model

If the phenomenom being studied is divided into separate events, which together cover the whole phenomenom, it is said that the sum of these probabilities = 1 (=100%). In the case of seperate events it is possible to talk about the addition theorem.

The overall situation is often depictable with a tree model, where different events are recorded with their respective probabilities. A tree model is read from top to bottom, so that one picks the branch that describes a series of events, and then multiply all the different odds of that branch. In the end, the probabilities of different branches are then summed.

Tree Diagram A graphical device used to list all possibilities of a sequence of events in a systematic way.

Example 6.There are three production lines in a factory, A, B and C. Of the total production 40% goes through line A, 25% through line B and the rest through line C. 3% of the products of line A, 1% of the products of line B and 2% of the products of line C are malfunctional. A random product is picked from a large batch. What are the odds that the product in question is malfunctional?



It is possible to calculate the answer directly from the model.P(the product is malfunctional) = 0.40 × 0.03+ 0.25 × 0.02 + 0.35 0.01 =

0.0205 º 2%

In practice we have come along the addition theorem as a part of the tree model's solution. When a problem includes different choices and options, then

- The consequent events "first A, then B" point to multiplication- The parallel events "A or B" point to addition

4.

It is expected that the following probabilities take place:1º If it doesn't rain on any chosen day , the probability for rain the following day

is Å1

3

2º If it rains on any chosen day, the probability for sunshine the following day is Å1

2What is the probability for rain the day after tomorrow, if it rains today?

Tree model! A= "Rain tomorrow, when it rains today"



Input @ P HAL =1

2×

1

3+

1

2×

1

2

5

12

4. Addition theorem

If the events A and B are independant, ie. when the other takes place the other cannot, there is a probability that the result of the random phenomenon is "A or B".

P(A or B) = P(A) + P(B)

Example 7.What is the probability for a two-figure number being divisible bya) two b) three c) six d) two or three?

Two-figure numbers: 10, 11, 12, ..., 99; 190 altogetherNumbers divisible by two: 10, 12, ..., 98; 45 altogetherNumbers divisible by three: 12, 15, ..., 99; 30 altogetherNumbers divisible by six: 12, 18, ..., 96; 15 altogether

a) P(divisible by 2) = 45

90= Å

1

2= 50%



b) P(divisible by 3) = 30

90= Å

1

3º 33.3%

c) P(divisible by 6) = 15

90= Å

1

6º 16.7%

d) Because numbers divisible by six are divisible by both two and three they are accountable in the amount of numbers divisible by two as well as three. In other words, they are involved twice in the addition 45+30. Thus, there are 45 + 30 - 15 numbers divisible by two or three.

P(divisible by 2 or 3) = 45 + 30 - 15

90 =

60

90 = Å

2

3 = 66.7%

Example 8.A sweets factory produces its "mix of the best"-assortment, of which 45% of the sweets contain additive A and 32% additive B. Both additives are found in 15% of the sweets. A candy is picked from a batch at random. What are the odds that the candy contains a) atleeast one of the two additives b) neither of the additives?

Now we cannot use figures the same way we did in example 6.

Let's make it easier to grasp the situation by tabulating the given information. Since both additives can be found in 15% of the sweets, the percentage of sweets containing only additive A is 45-15 = 30 and only additive B 32-15 = 17.

Additive APositive Negative

Additive B Positive 15 32-15=17Negative 45-15=30

a) P(atleast one additive) = 15% + 17% + 30% = 62%b) In the remaining section, the cell corresponding the lower right corner of the table, no additives have been used.

P(neither of the additives)= 100% - (15% + 17% + 30%) = 38% of the sweets.

In the case of examples 7 and 8 the given events were not independant, but there was the possibility of them taking place at the same time. In this case, like presented in the examples, the elimination of the events' redundancy is required while calculating the probability for 'A or B'.



5. Probability with a certain condition

It is often on purpose that we examine event A, in relation with a random event, in a smaller range of correspondance than the original population.

Example 9.

Let's think of the situation in example 6, where we calculated the probability for a product being malfunctional and we got the result

P(the product is malfunctional) = 0.205

If we are especially interested in the probability for the malfunctional product having been manufactured on production line A, it is natural to think of the group formed by the malfunctional products as a completely new domain.

The probability for a product being from line A while acknowledging that the product is malfunctional is found by examining the share of products from line A in comparison with the number of malfunctional products from all production lines.

P(the malfunctional product is from production line A) =

0.40 µ 0.03

0.0205 = 0.585... º 59%

In these type of situations we speak of conditional probability. We will not go further into the subject or using conditional probability on a wider scale.

5.

In a class of 35 all students study either French or German. 32 students study French and 29 study German. How many studya) bothb) only Frenchc) only German?



French = greyGerman = magenta

Answer a) 26 b) 6 c) 3

6.

Out of the two parking lots of a small store, both are taken 40 minutes an hour. 32 minutes of the time both lots are taken simultaneously. Two drivers arrive at the store at the same time. What are the odds that these two will find parking space immediately?



Input @

Answer : P H"Both get a lot" L =

1 − P H"atleast one of the lots is taken" L =

1 −8 + 32 + 8

60=

12

60=

1

5

7. What are the odds for a randomly chosen natural number to be divisible bya) the numbers 4, 5 and 6?b) the numbers 4, 5 or 6?

Every fourth natural number is divisible by 4.Every fifth number is divisible by 5 and every sixth by 6.Every twelth number is divisible by 4 and 6.Every twentieth number is divisible by 4 and 5.Every thirtieth number is divisible by 5 and 6.Every sixtieth number is divisible by 4, 5 and 6.

Input @ LCM@4, 5, 6 D60



Input @1

4+

1

5+

1

6−

1

12+

1

20+

1

30

9

20

Answer a) 1

60 b) Å

1

4+ Å

1

5+ Å

1

6-

1

12+

1

20+

1

30=

9

20

Geometrical probabilityOpen ê Close

If the sample space is possible to be viewed as a geometrical groups such as line segment (= length measure), sector (=angle or area measure), plane (= area measure) or space area (= volume measure), the probability for event A is P(A)=

The geometrical measure of A

The measure of whole sample space.

Example 1. Straight is drawn from point (1,3) to random direction. What is the probability that straight intersects the positive x-axis?

Input @

Clear @f, x D;

f @x_D = 3 x

g@x_D = 3

3 x

3

Input @ MDPlot @8f @xD, g @xD<, 8x , −1 , 3 < D



Input @

a

b

b

b

-1 1 2 3"x"

-2

2

4

6

8

"y"

Calculate angle a = 180°- b. Angle tan b = k (slope). The probability for

the event is the ratio P(A) = a

180.

Input @

Input @

Æ Pure NSolve ;

Clear @xD;

MDSelectSolution @MDRealOnly @NSolve @8 Tan@β Degree D == 3<, 8 β < DD, 1D

88β → 71.5651 <<Input @ β = 71.5651

71.5651

Input @ NumberForm@β, 4 D+71.57

Input @ α = 180 − β

Input @ 108.43494882292201



Input @ p HAL =108.43

180

0.602389

Answer: 0.60

Example 2. The total length of the Channel Tunnel is 50 kilometers and the

underwater part is 38 km. A train which is 400 meters long stops at random place in this railway between Calais and Dover. What is the probability that the train isa) entirelyb) entirely or partly under water?

http://fi.wikipedia.org/wiki/Tiedosto:Course_Channeltunnel_en.png

Solution: Focus on the front of the locomotive and find the favorable distance.a) The train is entirely under water if the front of the locomotive is not longer than 0,400 km from the rear end of the underwater tunnel.



Input @ P HAL =38 − 0.400

50

0.752

b) The train is entirely or partly under water if the front of the locomotive is inside the underwater tunnel or not more than 0,400 km past the end of the tunnel.

Input @ P HBL =38 + 0.400

50

0.768

Answer a) 0.752 b) 0.768

Example 3. The coefficients a and b of the straight ax - by = 0 are picked randomly from

between [0,1]. What is the probability that the slope of the straight is § Å1

3 ?

Solution:

The equation for the straight can be written to a form y = Åa

b x, where the slope Å

a

b can be

taken. So a

b§ Å

1

3, when b ¥ 3 a. The shadowed area at the picture corresponds that. The

solution can be calculated by calculating the ratio of the areas of the triangle and the square.



Input @

Input @P HAL =

1 × �13

2

1 × 1

�1

6

Answer: Å1

6

Example 4. The radius of the dartboard's bulsseye is 2 cm, score 9 is 4cm etc. The perimeter is 1cm wide. The dart hits random score. Calculate the expected value.

Solution: Calculate (!!) each score's area of the circle divided by the area of the whole board. At first calculate the area of the score "10", which is 4p.

So the area of the score "9" is p 42 - p 22 = 12p etc. Expected value = ‚i=1

n

xi ÿpi.



Input @

Distribution table

Input @

X p

10 π 22

π 212

9 π 42−π 22

π 212

8 π 62−π 42

π 212

7 π 82−π 62

π 212

6 π 102− π 82

π 212

5 π 122−π 102

π 212

4 π 142−π 122

π 212



3 π 162−π 142

π 212

2 π 182−π 162

π 212

1 π 202−π 182

π 212

0 π 212−π 202

π 212

Σ 1

Input @

Distribution = ::10,4

441>, :9,

4

147>,

:8,20

441>, :7,

4

63>, :6,

4

49>,

:5,44

441>, :4,

52

441>, :3,

20

147>,

:2,68

441>, :1,

76

441>, :0,

41

441>>

Input @

::10,4

441>, :9,

4

147>,

:8,20

441>, :7,

4

63>, :6,

4

49>,

:5,44

441>, :4,

52

441>, :3,

20

147>,

:2,68

441>, :1,

76

441>, :0,

41

441>>

Expected value =

10 ×4

441+ 9 ×

4

147+ 8 ×

20

441+



7 ×4

63+ 6 ×

4

49+ 5 ×

44

441+ 4 ×

52

441+

3 ×20

147+ 2 ×

68

441+ 1 ×

76

441+ 0 ×

41

441

Input @220

63êê N

3.49206

Answer: Expected value º 3 .5

8.

Solved Problem: Britney and David arrive independantly at the same bar between 19:00 and 20:00 in the evenings, staying in the bar for 15 minutes each time. What's the probability for the two being in the bar at the same time even for a moment?

Solution: X = "Britney's time of arrival" and Y = "David's time of arrival".The two will meet if X - Y § 15 and Y - X § 15 meaning Y ¥ X - 15 and Y § X + 15. We draw the lines Y = X - 15 and Y = X + 15. The favorable zone is the area lined out by the two lines.All possibilities are depicted by a 60x60 square.

Input @MDPlot @8x − 15, x + 15, 60 <, 8x , 0 , 60 <,

PlotRange → 88 0, 60 < , 80, 60 <<, Filling → 81 → 82<<D



Input @

Input @602 − 2 × 45×45

2

602

7

16

Answer: P("meet") = 7

16



Binomial probabilityOpen ê Close

Permutation An arrangement of objects in a specific order. In your Calculator, the button nPr.

If there are n elements in a set, the elements can be arranged in n! different orders. The arrangements are called permutations. From a set of n elements can be formed n*(n-1)*(n-2)* ... *(k-1) = n [nPr] permutations of k elements.

Example 1. Seven brothers (of Jukola) can arrange themselves in 7! = 5040 different orders.

Example 2. The brothers can elect a chairman, secretary and a treasurer in 7*6*5 = 7 [nPr 3] = 210 ways.

Combination A selection of objects without regard to order. In your Calculator, the button nCr.

If there are n elements in a set, you can form subsets which have k elements in

nk

= n!

Hn - kL! k! = n [nCr] k ways.

In Mathematica nk

= Binomial[n,k].



These subsets are called combinations.

Example 3. The seven brothers can elect 3 people amongst themselves to go to the store in

73

= 7!

H7 - 3L! 3!= 7 [nCr] 3 = 35 ways.

Example 4. In a mycology course the students were taught to idetify 78 different mushrooms, of which one of the students only learned 49. What is the probability that he would identify 6 randomly selected mushrooms.

Solution:

There are 786

= 256 851 595.different groups of 6

mushrooms.

There are 496

= 13 983 816. favorable cases.

The probability that all the mushrooms will be identified correctly is

496

786

= 0,054

Example 5. How big must a class be that it is favorable to bet (with even odds) that at least two students have the same birthday?

Solution:



The number of students = nP("At least two have the same birthday") = 1- P("all have different birthdays") =

1- 365

365.

364

365.363

365. . …

365 - Hn - 1L365

. > 0,50

365

365.

364

365.363

365. . …

365 - Hn - 1L365

< 0,50

Calculate the smallest value for n, so that 365 @nPrD n

365n < 0,50.

Table the values:

n365 @nPrD n

365n

10365 @nPrDµ 10

36510= 0,883

20 0,589

25 0,421

24 0,462

23 0,493

22 0,52



Answer: there must be at least 23 students.

Binomial ExperimentA binomial experiment is an experiment which satisfies these four conditions ⋅ A fixed number of trials ⋅ Each trial is independent of the others ⋅ There are only two outcomes ⋅ The probability of each outcome remains constant from trial to trial. These can be summarized as: An experiment with a fixed number of independent trials, each of which can only have two possible outcomes. The fact that each trial is independent actually means that the probabilities remain constant.

Examples of binomial experiments⋅ Tossing a coin 100 times to see how many heads occur. ⋅ Asking 500 people if they watch news. ⋅ Rolling a die to see if a 6 appears. Examples which aren't binomial experiments⋅ Rolling a die until a 6 appears (not a fixed number of trials) ⋅ Asking 20 people how old they are (not two outcomes) ⋅ Drawing 5 cards from a deck for a poker hand (done without replacement, so not independent)

Binomial Probability FunctionExample 6:What is the probability of rolling exactly two sixes in 6 rolls of a die? There are five things you need to do to work a binomial story



problem. Define Success first. Success must be for a single trial. Success = "Rolling a 6 on a single die" Define the probability of success (p): p = 1/6 Find the probability of failure: q = 5/6 Define the number of trials: n = 6 Define the number of successes out of those trials: k = 2 Anytime a six appears, it is a success (denoted S) and anytime something else appears, it is a failure (denoted F). The ways you can get exactly 2 successes in 6 trials are given below. The probability of each is written to the right of the way it could occur. Because the trials are independent, the probability of the event (all six dice) is the product of each probability of each outcome (die) 1 FFFFSS 5/6 * 5/6 * 5/6 * 5/6 * 1/6 * 1/6

= Å1

6

2 Å

5

6

4

2 FFFSFS 5/6 * 5/6 * 5/6 * 1/6 * 5/6 * 1/6

= Å1

6

2 Å

5

6

4

3 FFFSSF 5/6 * 5/6 * 5/6 * 1/6 * 1/6 * 5/6

= Å1

6

2 Å

5

6

4

4 FFSFFS 5/6 * 5/6 * 1/6 * 5/6 * 5/6 * 1/6

= Å1

6

2 Å

5

6

4

5 FFSFSF 5/6 * 5/6 * 1/6 * 5/6 * 1/6 * 5/6

= Å1

6

2 Å

5

6

4

6 FFSSFF 5/6 * 5/6 * 1/6 * 1/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

7 FSFFFS 5/6 * 1/6 * 5/6 * 5/6 * 5/6 * 1/6



= Å1

6

2 Å

5

6

4

8 FSFFSF 5/6 * 1/6 * 5/6 * 5/6 * 1/6 * 5/6

= Å1

6

2 Å

5

6

4

9 FSFSFF 5/6 * 1/6 * 5/6 * 1/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

10 FSSFFF 5/6 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

11 SFFFFS 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 1/6

= Å1

6

2 Å

5

6

4

12 SFFFSF 1/6 * 5/6 * 5/6 * 5/6 * 1/6 * 5/6

= Å1

6

2 Å

5

6

4

13 SFFSFF 1/6 * 5/6 * 5/6 * 1/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

14 SFSFFF 1/6 * 5/6 * 1/6 * 5/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

15 SSFFFF 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6

= Å1

6

2 Å

5

6

4

Notice that each of the 15 probabilities are exactly the

same: Å1

6

2 Å

5

6

4.

Also, note that the Å1

6 is the probability of success and you



needed 2 successes. The Å5

6 is the probability of failure, and if

2 of the 6 trials were success, then 4 of the 6 must be failures. Note that 2 is the value of x and 4 is the value of n-k. Further note that there are fifteen ways this can occur. This is the number of ways 2 successes can be occur in 6 trials without repetition and order not being important, or a combination of 6 things, 2 at a time. The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:

P(X=k) = nk

pk qn-k where q = 1-p (or P(X=k) = nCk

* pk * qn-k )Example 7:A coin is tossed 10 times. What is the probability that exactly 6 heads will occur. Success = "A head is flipped on a single coin" p = 0.5 q = 0.5 n = 10 x = 6

P(X=6) =106

o .56 o .510-6 (or 10C6 * 0.5^6 * 0.5^4) = 210 *

0.015625 * 0.0625 = 0.205078125

Mean, Variance, and Standard DeviationThe mean, variance, and standard deviation of a binomial distribution are extremely easy to find. m = np

d2 = npq

d = npq



Example 8.Find the mean, variance, and standard deviation for the number of sixes that appear when rolling 30 dice. Success = "a six is rolled on a single die". p = 1/6, q = 5/6. The mean is 30 * (1/6) = 5. The variance is 30 * (1/6) * (5/6) = 25/6. The standard deviation is the square root of the variance = 2.041241452 (approx)

Example 9. When a coin is thrown five times. What are the changes of getting second heads with the last throw?The heads has to occur exactly one time in the 4 first throws.

P("heads occurs second time in the last

throw")=41

Å1

2

1Å1

2

3Å1

2= Å

1

8

Answer: The changes are Å1

8.

In repeating the same random trial n times we receive the probability for event A happening exactly k times

P(Ak) =nk

pk qn-k , where the probability for a single

event A occuring is P(A) = p , and q = 1 - p.

9. A) What is the probability for getting atleast two sixes on five dice rolls?B) How many dice rolls are necessary for the probability for getting atleast one six to rise over 90?

A) Solution:By complementary event.P("Atleast two sixes") = 1 - P("One six at most")

n = 5 ;



p =1

6;

x = 0 ;

PDF@BinomialDistribution @n, p D, x D êê N

H∗ PH X = x L = ? ∗L0.401878

Input @ p0 = 0.4018775720164609

0.401878

Input @

n = 5 ;

p =1

6;

x = 1 ;

PDF@BinomialDistribution @n, p D, x D êê N

H∗ PH X = x L = ? ∗L0.401878

Input @ p1 = 0.4018775720164609

0.401878

Input @ 1 − Hp0 + p1L0.196245

Answer: 0.196 (=19.6)

B) the number of necessary throws=n

Input @

x1 = 1 ;

probability = 0.90 ;

MDSFindnBinomialDistributionPnp @ x1 ≤ X ≤ n,

8p, probability <DH∗ find n that P Hx1 ≤ X ≤ nL = probability ∗L



0 1 2 3 4 5 6 7 8 9 10 11 12 13

0.05

0.10

0.15

0.20

0.25

0.30Probability

PH1 § X § 13L º0.91

Binomial Distribution

n p µ σ PH 1 ≤ X ≤ 13L13 �1

62.16667 1.34371 0.90653612101282076311

Answer: Atleast 13 throws.

10.In the second round of the 2000 Finnish presidential elections Tarja Halonen received 51,6% of the votes whereas Esko Aho received 48,4%. What is the probability for a random group of five voters voting for Esko Aho?

Input @

n = 5 ;

p = 0.484 ;

x = 3 ;

MDSBinomialDistributionPnp @ x ≤ X , 8n, p <DH∗ PH value ≤ X L = ? ∗L



0 1 2 3 4 5

0.05

0.10

0.15

0.20

0.25

0.30

ProbabilityPH3 § XL º0.47


n p µ σ PH 3 ≤ XL5 0.484 2.42 1.11746 0.47002047370854400000

Answer: 0.47

11.The probability for a certain seed type to sprout is 75%. How many seeds do you have to plant in order to gain the probability for atleast one plant sprouting over 99,9%?

Input @

p = .75 ;

x1 = 1 ;


MDSFindnBinomialDistributionPnp @ x1 ≤ X ≤ n,

8p, probability <DH∗ find n that P Hx1 ≤ X ≤ nL = probability ∗L



0 1 2 3 4 5

0.1

0.2

0.3

0.4

ProbabilityPH1 § X § 5L º0.9990


n p µ σ PH 1 ≤ X ≤ 5L5 0.75 3.75 0.968246 0.99902343750000000000

or alternatively

Input @

p = 0.75 ;

x1 = 1 ;


8nstart = 1, nend = 6<; H∗ change nstart,nend ∗L

MDSFindnBinomialDistributionSteps @ x1 ≤ X ≤ n,

8p, probability <, 8nstart, nend <DH∗ find n that P Hx1 ≤ X ≤ nL = probability ∗L

n PH x1 <= X <= n L Wanted P1 0.75000000000000000000 0.999

2 0.93750000000000000000 0.999

3 0.98437500000000000000 0.999

4 0.99609375000000000000 0.999

5 0.99902343750000000000 0.999

6 0.99975585937500000000 0.999



1 2 3 4 5 6n

0.2

0.4

0.6

0.8

1.0

PH x1 <= X <= n L

Answer: Atleast 5 seeds.



Multinomial ProbabilitiesOpen ê Close

Multinomial ProbabilitiesA multinomial experiment is an extended binomial probability. The difference is that in a multinomial experiment, there are more than two possible outcomes. However, there are still a fixed number of independent trials, and the probability of each outcome must remain constant from trial to trial. Instead of using a combination, as in the case of the binomial probability, the number of ways the outcomes can occur is done using distinguishable permutations. An example here will be much more useful than a formula. The probability that a person will pass a College Algebra class is 0.55, the probability that a person will withdraw before the class is completed is 0.40, and the probability that a person will fail the class is 0.05. Find the probability that in a class of 30 students, exactly 16 pass, 12 withdraw, and 2 fail.

Outcome x p Ioutcome LPass 16 0.55

Withdraw 12 0.40

Fail 2 0.05

Total 30 1.00

The probability is found using this formula:

P =30!

H16!L H12!L H2!L * 0.5516* 0.4012 * 0.052 (Count this ;) )

12.

The odds for a serial producted single-use item having one error is 7%, two errors is 3% and no errors is 90%. What is the probability in a set of hundred items a) there are exactly 90 products without errors, 7 products with 1 error and 3 products with 2 errors,b) at least one product having at least one error?



Hypergeometric ProbabilitiesOpen ê Close

Hypergeometric ProbabilitiesHypergeometric experiments occur when the trials are not independent of each other and occur due to sampling without replacement -- as in a five card poker hand (see the Solved Problem 5.5 later). Hypergeometric probabilities involve the multiplication of two combinations together and then division by the total number of combinations.

Example 1.How many ways can 3 men and 4 women be selected from a group of 7 men and 10 women?

The answer is =

73

104

177

= 7350/19448 º 0.3779

Note that the sum of the numbers in the numerator are the numbers used in the combination in the denominator. This can be extended to more than two groups and called an extended hypergeometric problem.

Five cards are dealt from the deck.What is the probability that you geta) a pair b) two pairsc) three of a kindd) straight e) flushf) full houseg) four of a kindh) straight flush?

Solution:13 4 12 4 4 4



a)P(”a pair”) =

131

42

123

41

41

41

525

nk

= Binomial[n,k]

b) P(”two pairs”) =

132

42

42

111

41

525

c) P(”three of a kind”) =

131

43

122

41

41

525

d) Note. also includes a straight flush P(”straight”) = 10 * 45

525

e) Note. also includes a straight flush P(”flush”) = 41

135

525

f ) P(”full house”) =

131

43

121

42

525

g) P(”four of a kind”) =

131

44

121

41

525

h) P(”straight flush”) = 4 * 10

525

a)

Input @

HBinomial @13, 1 D ∗ Binomial @4, 2 D ∗ Binomial @12, 3 D ∗ Binomial @4, 1 D ∗

Binomial @4, 1 D ∗ Binomial @4, 1 DL ∗1

Binomial @52, 5 D



Input @352

833êê N

0.422569

b)

Input @

HBinomial @13, 2 D ∗ Binomial @4, 2 D ∗ Binomial @4, 2 D ∗


Binomial @52, 5 D

Input @198

4165êê N

0.047539

c)

Input @

HBinomial @13, 1 D ∗ Binomial @4, 3 D ∗ Binomial @12, 2 D ∗


Binomial @52, 5 D

Input @88

4165êê N

0.0211285

d)

Input @10 ∗ 45

Binomial @52, 5 D

Input @128

32 487êê N

0.00394004

e)

Input @ HBinomial @4, 1 D ∗ Binomial @13, 5 DL ∗1

Binomial @52, 5 D

Input @33

16 660êê N

0.00198079

f)

Input @

HBinomial @13, 1 D ∗ Binomial @4, 3 D ∗


Binomial @52, 5 D

Input @6

4165êê N



0.00144058

g)

Input @

HBinomial @13, 1 D ∗ Binomial @4, 4 D ∗


Binomial @52, 5 D

Input @1

4165êê N

0.000240096

h)

Input @10 ∗ 4

Binomial @52, 5 D

Input @1

64 974êê N

0.0000153908

Discrete DistributionsOpen êClose

Random variable's x value is ruled by chance.-A random variable is discrete or in other words discontinious, if it can get only exact values.-A random variable is continuous, if it can get any values.

Example 1. Define a random variable x's="the sum of two die rolls" probability distribution and cumulative distribution function. Calculate the expected value and the standard deviation of the distribution. Present the distribution graphically.

x Calc . Prob. Cumul .2 Å1

6* Å1

6136

136

3 Å26

* Å16

236

336

4 Å36

* Å16

336

636

5 Å46

* Å16

Å46

1036

6 Å56

* Å16

536

1536

7 Å66

* Å16

636

2136



8 Å56

* Å16

536

2636

9 Å46

* Å16

436

3036

10 Å36

* Å16

336

3336

11 Å26

* Å16

236

3536

12 Å16

* Å16

Å16

3636

S - 1 1

The expected value m = 2 *1

36+ 3 *

2

36+4*

3

36...= 7

s=

Si-1

n Jxi - x

-Nn - 1

=H2 - 7L2 + 2 * H3 - 7L2 + 3 * H4 - 7L2 + ... + H12 - 7L2

35º 4.971

14.Five cards are drawn from a deck of cards. Form a distribution for a random variable x, which tells the number of the cards from the most frequently drawn suit.

x-

p ~p

2 B 41

132

33

131

131

131

+41

132

32

132

131

Fì 525

~62.89 %

3 B 41

133

32

131

131

+41

133

31

132

Fì 525

~32.62 %

4 B 41

134

31

131

Fì 525

F ~4.29 %

541

135

ì 525

~0.198 %

⁄ - 1

Binomial distributionOpen êClose

x~Bin(n,p)Expected value m = np

Standard deviation s = npq

Example 1. A die is rolled five times. Form a distribution for the random variable x that tells the



frequency of sixes. Calculate the expected value and the standard deviation. Display the distribution graphically.

Input @

n = 5 ;

p =1

6;

MDSBinomialDistributionPlot @8n, p <,

MD2σRegion → True D

0 1 2 3 4 5X

0.1

0.2

0.3

0.4

Probability

Binomial Distribution with n = 5, p = 0.166667


µ σ P Hµ − 2 σ ≤ X ≤ µ + 2 σL0.833333 0.833333 0.964506

Answer: Both the expected value and the standard deviation are Å5

6

Example 2.Approximately 8% of Finnish men suffer from color blindness. What is the probability that out of schools 400 boys...

a) 2 or less suffer from color blindness

b) atleast two suffer from color blindness

n = 400 ;

p = 0.08 ;



x = 2 ;

MDSBinomialDistributionPnp @ X ≤ x , 8n, p <DH∗ PH X ≤ value L = ? ∗L

99 199 299 399

0.01

0.02

0.03

0.04

0.05

0.06

0.07

ProbabilityPHX § 2L RowBox@8<, , SuperscriptBox@10, RowBox@8-, 9<DD<D


n p µ σ PH X ≤ 2L400 0.08 32. 5.42586 2.0929443760062278182 × 10−12

Input @

n = 400 ;

p = 0.08 ;

x = 3 ;

MDSBinomialDistributionPnp @ x ≤ X , 8n, p <DH∗ PH value ≤ X L = ? ∗L



99 199 299 399

0.01

0.02

0.03

0.04

0.05

0.06

0.07

ProbabilityPH3 § XL > 0.9999


n p µ σ PH 3 ≤ XL400 0.08 32. 5.42586 0.99999999999790705562

Answer: The probability of having 2 or less suffering from color blindness is around 0. Atleast two suffer from color blindness is around 1.

Poisson distributionOpen êClose

Named after the French mathematician Simeon Poisson, Poisson probabilities are useful when there are a large number of independent trials with a small probability of success on a single trial and the variables occur over a period of time. It can also be used when a density of items is distributed over a given area or volume.

In a repeated trial the Poissson distribution is used instead of the binomial distribution,if -the exact values of the parameters n and p are not known-the number of trials is great and the probability of the examined event A is small



x ~ Poisson(a), a is the expected value

P(x = k) =ak

k ! e-a

If there are 500 customers per eight-hour day in a check-out lane, what is the probability that there will be exactly 3 in line during any five-minute period? The expected value during any one five minute period would be 500 / 96 = 5.2083333. The 96 is because there are 96 five-minute periods in eight hours. So, you expect about 5.2 customers in 5 minutes and want to know the probability of getting exactly 3.

P(X = 3) =

K 500

96O3

3!e

-50096

Input @

λ = 3;

x = 3;

N@PDF@PoissonDistribution @λD, x DDH∗ PHX = xL=? ∗L

0.224042

Example 1.The average European Football Championship final tournament game witnessed 2,5 goals. What was the probability that in a randomly picked game...

a)three goals were made?

b)atleast 2 goals were made?

a)P(x = 3) = 2.53

3! ‰ -2.5

Input @

λ = 2.5 ;

x = 3;

N@PDF@PoissonDistribution @λD, x DDH∗ PHX = xL=? ∗L

0.213763

b)P("atleast two goals")=1- P("one or no goals")=1-[2.50

0! ‰ -2.5+

2.51

1! ‰ -2.5]º0.71



λ = 2.5 ;

x = 2 ;

MDSPoissonDistributionP @ x ≤ X , 8λ<DH∗ PH value ≤ X L = ? ∗L

1 3 5 7 9 11 13X

0.05

0.10

0.15

0.20

0.25


Poisson Distribution

λ µ σ PH 2 ≤ XL2.5 2.5 1.58114 0.71270250481635421691

Answer: The probability of 3 goals was 0.21 The probability of atleast two goals was 0.71

Example 2Merja gets on average 20 text messages a day (12h). What is the probability of Merja getting atleast three text messages during the next hour.

P("3 or more text messages") = 1 - P("two or less goals") =

1 - [J 20

12N0

0!‰

-2012 +

J 20

12N1

1! ‰

-2012 +

J 20

12N2

2! ‰

-2012 ]



Input @

λ =20

12;

x = 3 ;


1 3 5 7 9X

0.05

0.10

0.15

0.20

0.25

0.30



λ µ σ PH 3 ≤ XL1.66667 1.66667 1.29099 0.23400449960322143291

Answer: The probability of Merja getting atleast 3 text messages during the next hour is 23.4%

15.The number of calls to a call center follows the Poisson distribution. Calculate the probability that the call center receives atleast 5 calls a minute, when the center's call frequency a is 3.

P("atleast 5 calls a minute")=1-P("4 or less calls a minute") =

1 - [30

0!‰ -3+

31

1! ‰ -3+

32

2!‰ -3+

33

3!‰ -3+

34

4!‰ -3]



Input @

λ = 3 ;

x = 5 ;


1 3 5 7 9 11 13 15X

0.05

0.10

0.15

0.20



λ µ σ PH 5 ≤ XL3. 3. 1.73205 0.18473675547622793371

Answer: The probability of the call center getting atleast 5 calls a minute is 18%

16.

A new hamburger bar is designed to serve four customers a minute. During the lunch hour of the first day the bar received three customers a minute on average. Approximate the probability of demand exeeding the service capasity during the lunch hour of the next day.

a = 3

The demand exeeds the service capasity, if the bar gets more than 4 customers a minute.

P("more than 4 customers") = 1 - P("4 or less customers") =

1 - [30

0! ‰ -3+

31

1!‰ -3+

32

2!‰ -3+

33

3!‰ -3 +

34

4! ‰ -3]



λ = 3 ;

x = 5 ;


1 3 5 7 9 11 13 15X

0.05

0.10

0.15

0.20



λ µ σ PH 5 ≤ XL3. 3. 1.73205 0.18473675547622793371

Answer: The probability of demand exeeding service capacity is 18%



Continuous DistributionsOpen êClose

Example 1.

Busses arrive to a bus stop in 15 min intervals. A person who doesn't know the bus schedule arrives to a stop at a random time. What is the probability that...

a)... the person has to wait for a maximum of five minutes

b)...the person hast to wait for a mimimum of eigth minutes

c)...3-6 min

d)exactly 10 minutes

e)What is the expected value of the wait?

a) 5

15 = Å

1

3

b) 7

15

c) 3

15 = Å

1

5d) 0e) 7.5 min

17.By what value of the constant b, f(x) = 1 - Å

1

8x, when 0 § x § b, otherwise 0, is a

probability density function?Then calculate P(x ¥ 1)

A = 1

Input @ MDPlot B:1 −1

8 x, >, 8x , 0 , 8 < F



2 4 6 8x

0.2

0.4

0.6

0.8

1

y

1 * K1 - Å18

bO2

*b = 1

-1

16b2 + Å

1

2b - 1 = 0

Input @

Æ Pure Solve ;

Clear @xD;

MDRealOnly BSolve B: −1

16 b2 +

1

2 b − 1 == 0>, 8 b < FF

88b → 4<, 8b → 4<<

Answer: b is 4.

18.

The side length of a square shaped room is 4m. The random variable x is the distanse of a glass bead dropped on the room surface from the closest wall.

a) Define P(x § 0.5)

b) Define the random variable's cumulative distribution function and plot it

c) Define the random variable's probability density function and plot it

a)



The area of the gray part of the room is 42- H4 - 2 xL2 = 16 - H4 - 2 xL2.

The probability that x § 0.5 is 16 - H4 - 2 * 0.5L2

16

Input @16 − H4 − 2 ∗ 0.5 L2

16

0.4375

b) The cumulative distribution function is I16 - H4 - 2 xL2M

16 when 0 § x § 2, 1 when x > 2,

and 0 when x < 0.

Input @ Clear @xD;

Input @ F@x_D = I16 − H4 − 2 xL2M ë 16

1

16I16 − H4 − 2 xL2M

Input @ MDPlot @8F@xD<, 8x , 0 , 2 < D



0.5 1 1.5 2x

0.2

0.4

0.6

0.8

1

y

� Graphics �

c) The density function is the derivative of the cumulative function.

Input @ F' @xD1

4H4 − 2 xL

when 0 § x § 2, and 0 elsewhere.

Input @ MDPlot @8F ' @xD<, 8x , 0, 2 < D

0.5 1 1.5 2x

0.2

0.4

0.6

0.8

1

y

� Graphics �



Normal distributionOpen êClose

x ~ N(m,s)

The expected value is m and the standard deviation is s.

Z = x - m

s

Example 1

Intelligence quotient is measured with a test, which average is 100 and standard deviation is 15.What is the probability that a randomly picked person has an IQ of under 97?

The IQ limit 97 correlates to norm value 97 - 100

15= -0.2

P(x § -0.2) = F(-0.2) = 1 - F(0.2) = 0.42074

Input @

Æ ;

µ = 100 ; σ = 15 ; x = 97 ;

probability = MDSφ@x, 8µ, σ<D;

MDSNormalDistributionP µσ@ X ≤ x , 8µ, σ<DH∗ PH X ≤ value L = ? ∗L



40 60 80 120 140 160X

PHX § 97L = 0.42074

Normal Distribution

z H−∞L z H97. L PH X ≤ 97. L−∞ −0.2 0.42074

Dev Hz σL100. −3.

Answer: The probability of a randomly picked person scoring under 97 in this test is 42%.

19.The weight of boys at the age of 18 follows the Normal distribution, so that the average weight is 66 kg and the standard deviation is 8kg. What percentage of 18 year old boys weigh from 60 to 70 kg?

P("between 60kg and 70kg") = P("under 70kg") - P("under 60kg").

Input @

µ = 66 ; σ = 8 ; x1 = 60 ; x2 = 70 ;

probability = MDSφ@x2, 8µ, σ<D − MDSφ@x1, 8µ, σ<D;

MDSNormalDistributionP µσ@ x1 ≤ X ≤ x2 , 8µ, σ<DH∗ PH x1 ≤ X ≤ x2 L = ? ∗L



30 40 50 60 70 80 90 100X

PH60 § X § 70L = 0.464835

Normal Distribution

z H60. L z H70. L PH 60 ≤ X ≤ 70L−0.75 0.5 0.464835

Answer: 46% of boys weigh between 60kg and 70kg.

20.

Let's presume that the mass per area in paper is normally distributed. A client orders paper which mass per area ratio is 80 gím2. How great can the mass

per area ratio's deviation be for the probability of getting paper with less than 75 gím2 mass per area is less than 5%?

z = x - m

s, s =

x - m

z

Input @


Clear @zD; z = z ê. FindRoot @MDSφ@zD == probability, 8z, 0 <D;

MDSNormalDistributionP µσ@ Z ≤ z, 80, 1 <DH∗ PH Z ≤ ? L = probab ∗L



-4 -2 2 4Z

PHZ § -1.64485L = 0.05

Normal Distribution

z H−∞L z H−1.64485 L PH Z ≤ −1.64485 L−∞ −1.64485 0.05

Dev Hz σL0. −1.64485

Input @ σ =75 − 80

−1.64485

3.03979

Answer: The standard deviation can be 3.04 g/m2 at maximum.

21.One can get in to a society of top intelligent people, if one's intelligence is greater than that of 98% of the population. What IQ does one have to have to get into this society? IQ~(100,24)

The z of 98%

Input @






-4 -2 2 4Z

P HZ § 2.05375L = 0.98

-4 -2 2 4Z

P HZ § 2.05375L = 0.98

Normal Distribution

z H−∞L z H2.05375 L P H Z ≤ 2.05375 L−∞ 2.05375 0.98

Dev Hz σL0. +2.05375

Input @ µ = 100

100

Input @ σ = 24

24

Input @ x = z ∗ σ + µ

149.29

Answer: For one to get in one has to have an IQ over 149.

22.

If xi(i = 1,2,3,... ,n) are normally distributed random variable's, which averages

are m i and standard deviations are s i, so likewise x1+x2+x3+...+ xn are N(m,s 2)

where m=m 1+m 2+m 3+...m n and s 2 =s 21+s 2

2+s 23+...+s 2

n

Solve the following problem: There are 23 sweets in the bag. The mass of the sweeties is normaly distributed, so that the average is 2,1g and the standard deviation is 0,25g. What is the probability that the sweets in the bag weigh in together at over 50g?

Input

Æ ;

µ = 23 ∗ 2.1 ; σ = 23 ∗ 0.25 2 ; X = 50 ;



23 2.1 ; 23 0.25 ; X 50 ;

z =X − µ

σêê N

1.18261

Input @

Æ ;

z = 1.1826086956521709` ;

probability = 1 − MDSφ@zD;

MDSNormalDistributionP µσ@ z ≤ Z, 80, 1 <DH∗ PH z ≤ Z L = ? ∗L

-4 -2 2 4Z

P H1.18261 § ZL = 0.118482

-4 -2 2 4Z

P H1.18261 § ZL = 0.118482

Normal Distribution

z H1.18261 L z H+∞L P H 1.18261 ≤ ZL1.18261 +∞ 0.118482

Dev Hz σL0. +1.18261

Answer: The probability of the bags weight topping 50 g is 11.8%.

23.

The mean velocity of Finnish traffic flow is 81 km/h with in 80 km/h area. 13.4% of drivers speed 10km/h or more over the allowed 80 km/h limit.

a) Calculate the standard deviation of velocities, when the velocities are presumed to be normaly distributed.

b) How great portion of the drivers don't survive with a traffic fine, but get a penalty demand in other words speed over 21 km/h?

a) Now P(x § 90) = 1 - 0.134 = 0.866






-4 -2 2 4Z

PHZ § 1.10768L = 0.866

Normal Distribution

z H−∞L z H1.10768 L PH Z ≤ 1.10768 L−∞ 1.10768 0.866

Dev Hz σL0. +1.10768

Input @ σ =90 − 81

1.10768

8.12509

b) We solve the standardized value of speeding 21km/h. (Driving at 101 km/h)

z = x - m

s

Input @ z =101 − 81

8.125

2.46154

Input @

Æ ;

z = 2.4615384615384617` ;




probability 1 MDS@zD;


-4 -2 2 4Z

P H2.46154 § ZL = 0.00691713

-4 -2 2 4Z

P H2.46154 § ZL = 0.00691713

Normal Distribution

z H2.46154 L z H+∞L P H 2.46154 ≤ ZL2.46154 +∞ 0.00691713

Dev Hz σL0. +2.46154

Answer: The standard deviation is 8.1 km/h0.69% of the drivers speed over 21 km/h.

Replacing the binomial distribution with the normal distributionOpen êClose

When number of repeats n is great the binomial distribution Bin(n,p) can be replaced with the normal distribution N(m,s), where m = np and

s = npq

Example 1.The probability of a customer returning a product he/she has ordered from mailorder is 11%. What is the probability, that out of mailorder company's 2500 customers 300 or more return the products they've ordered?



Input @

n = 2500 ;

p = 0.11 ;

:µ = n p , σ = n p H1 − pL > êê N

8275., 15.6445 <

Let's standardize 300,

Input @

Æ ;

µ = 275 ; σ = 15.644487847162015 ; X = 300 ;

z =X − µ

σêê N

1.59801

Input @

Æ ;

z = 1.59801 ;



-4 -2 2 4Z

P H1.59801 § ZL = 0.0550204

-4 -2 2 4Z

P H1.59801 § ZL = 0.0550204

Normal Distribution

z H1.59801 L z H+∞L P H 1.59801 ≤ ZL1.59801 +∞ 0.0550204

Dev Hz σL0. +1.59801

Answer: The probability of 300 or more customers returning their product is 5.5%



About 5% of people are left handed. What is the probability, that there are atleast 30 "lefties" in Lyseonpuiston lukio (Lyska) ? (circa 800 students)

Input @

n = 800 ;

p = 0.05 ;

:µ = n p , σ = n p H1 − pL > êê N

840., 6.16441 <

Input @

Æ ;

µ = 40 ; σ = 6.164414002968976` ; X = 30 ;

z =X − µ

σêê N

−1.62221

Input @

Æ ;

z = −1.6222142113076254` ;



-4 -2 2 4Z

P H-1.62221 § ZL = 0.947621

-4 -2 2 4Z

P H-1.62221 § ZL = 0.947621

Normal Distribution

z H−1.62221 L z H+∞L P H −1.62221 ≤ ZL−1.62221 +∞ 0.947621

Dev Hz σL0. −1.62221



Answer: The probability of 'Lyska' having atleast 30 "lefties" is 94.8%.

25.

A presidential candidate is supported by 48,4% of the voters. Calculate the probability, that the majority out of randomly picked group of 1000 people participating in an election poll are going to support our covered candite?

Input @

n = 1000 ;

p = 0.484 ;

:µ = n p , σ = n p H1 − pL > êê N

8484., 15.8033 <

Input @

Æ ;

µ = 484.` ; σ = 15.803290796539814` ; X = 500;

z =X − µ

σêê N

1.01245

Input @

Æ ;

z = 1.0124473570721901` ;



-4 -2 2 4Z

P H1.01245 § ZL = 0.155662

-4 -2 2 4Z

P H1.01245 § ZL = 0.155662



Normal Distribution

z H1.01245 L z H+∞L P H 1.01245 ≤ ZL1.01245 +∞ 0.155662

Dev Hz σL0. +1.01245

Answer: The probability of the candidate getting the majority in the poll is 15.6%.

Sample selectionOpen êClose

An avarege calculated from a sample is an estimate for the real distributions average i.e expected value. When we pick multiple same size samples the same variable in mind and calculate the average of each sample, results are very often normaly distributed. Even when the variable itself is not normaly distributed. In this kind of situation we use the average calculated from the sample as the expected value x and

as standard deviation we use the average's mean errors

n, where n is the size of the

sample. X and s are indexes of the random variable's distribution or if they are not known like the case often is, they are indexes calculated from the sample.

Example 1.The average weight of salmon fished out from the river Teno was 8,5 kg and the salmon's weight's deviation was 3 kg.What is the probability that...

a) the next fish weighs over 11kg?

b) the average weight of the next 20 fishes is over 10 kg?

a)

Input @

Æ ;

µ = 8.5 ; σ = 3 ; x = 11 ;

probability = 1 − MDSφ@x, 8µ, σ<D;

MDSNormalDistributionP µσ@ x ≤ X, 8µ, σ<DH∗ PH value ≤ X L = ? ∗L



-5 0 5 10 15 20X

P H11 § XL = 0.202328

-5 0 5 10 15 20X

P H11 § XL = 0.202328

Normal Distribution

z H11. L z H+∞L P H 11. ≤ XL0.833333 +∞ 0.202328

Dev Hz σL8.5 +2.5

b)

Input @

Æ ;

µ = 8.5 ; σ =3

20; x = 10 ;





6 7 8 9 10 11X

P H10 § XL = 0.0126737

6 7 8 9 10 11X

P H10 § XL = 0.0126737

Normal Distribution

z H10. L z H+∞L P H 10. ≤ XL2.23607 +∞ 0.0126737

Dev Hz σL8.5 +1.5

Answer: The probability of the next fish weighing over 11kg is 20.2%.The probability of the average of the next 20 being over 10 kg is 1.3%.

26.The average service time in a hamburger bar is 8 minutes and the standard deviation is 3 minutes. The bar has ten customers. Calculate the probability that the average service time for these customers exeeds 10 minutes.

Input @

Æ ;

µ = 8 ; σ =3

10; x = 10;





4 6 10 12X

P H10 § XL = 0.0175075

4 6 10 12X

P H10 § XL = 0.0175075

Normal Distribution

z H10. L z H+∞L P H 10. ≤ XL2.10819 +∞ 0.0175075

Dev Hz σL8. +2.

Answer: the probability is 1.8%



SummaryOpen êClose Print

Mathematica project summary, what have we learnt:

During this project we have learnt how to use Mathematica program and to cooperate with students from other countries from all around Europe. One of most notable gains we have also gotten from the project was learning basic mathematical English. This was due to numerous social interactions with students and teachers from different cultural backgrounds. Interaction between different cultures taught us also valuable lessons from other than mathematical fields.

Algebra group thought that the most valuable usage of the Mathematica and M@th Desktop is in graph drawing, it enables to imagine problems in mathematics better.

Functional group emphasizes that the most remarkable aspect of Mathematica and M@th Desktop is that it concretizes mathematics. For example by drawing the functional exercise it was easier to perceive the problem. The program was very flexible and it made it possible to solve difficult exercises in short time.

Students in trigonometry group have learnt many things about how to use Mathematica and M@th Desktop program. The group discovered that the program is a useful tool to solve many different types of trigonometrical exercises.

Probability group focused on probability in which they improved a lot during the project. The group learned of the several useful and efficient features of Mathematica and M@th Desktop for probability problems.

E-learning was one of the main focuses of the project and it was easy and still effective to solve difficult exercises with international teamwork. All groups think that the best part of the project was the necessity to communicate with students from foreign countries, especially including language practice, which still continues. Most of us have kept contact with each other also outside the project and we have made friends with them. Mathematica project has created a new aspect to our mathematical study techniques.



Our TeamOpen êClose Print

Gymnazium Teplice, Teplice, Czech RepublicJakub Krmela Jan MoravecZuzana Jaklinova Eva TrubenekrovaDavid Kubon Jan HodekJan Kubalik



Lyseopuistonlukio, Rovaniemi, FinlandTiina Alaruikka Miika Haataja Petri Huttu Leena HytönenIida Impiö Henri Izadi Annemari Kiviniemi Sami Kojola Essi-Riina Korva Laura Lehtiniemi Anna Maijala Johan Matinmikko Merja Mettänen Kalle Niemi Eero Pihkala Tanja Pohjola Antti Pyykkönen Laura Sandqvist Paulus Sieppi Noora Siitonen Tommi Tolonen Riina Typpö



The ZST Technical School, Mikolow, Poland



Gymnazium A. Bernolaka, Namestovo, Slovakia

New Chapter Cut Last Chapter

Open êClose Print

List of our sources:

We have taken some parts of the Linear function, Quadratic Functions and Trigonometric Functions notebook of the e-learning software M@th Desktop 5.0 and edited it.M@th Desktop 5.0 is based on Mathematica.

PCMLogo: PCM Homepage

Article: lllll

Formula: Booktitle, Author, Publ. Comp, Year, jméno knihy dopsat !!!



Documents

Math E-learning Materials - Lyskalyska.net/MathProject/Projects/Math e-learning/Math E-learing Materi… · Math E-learning Materials Gymnazium Teplice, Teplice, Czech Republic Lyseopuistonlukio,