Upload
clement-oliver
View
225
Download
0
Tags:
Embed Size (px)
Citation preview
Pre-AlgebraPre-Algebra
Objectives:1. To round decimals2. To estimate sums and differences
Rounding and EstimatingRounding and Estimating
Lesson 3-1
Pre-AlgebraPre-Algebra
Tip: ≈ means approximately equal to
Rounding and EstimatingRounding and Estimating
Lesson 3-1
Pre-AlgebraPre-Algebra
Rounding and EstimatingRounding and Estimating
Lesson 3-1
a. Round 8.7398 to the nearest tenth.
8.7398
tenths place
less than 5
8.7
Round down to 7.
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Rounding and EstimatingRounding and Estimating
Lesson 3-1
b. Round 8.7398 to the nearest integer.
8.7398
nearest integer is ones place
5 or greater
9
Round up to 9.
Additional Examples
Pre-AlgebraPre-Algebra
Estimate to find whether each answer is reasonable.
Rounding and EstimatingRounding and Estimating
Lesson 3-1
a. Calculation
+$ 59.98
$ 83.21
$115.67
$258.86+$ 60
$ 80
$120
Estimate
$260
The answer is close to the estimate. It is reasonable.
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Rounding and EstimatingRounding and Estimating
Lesson 3-1
b. Calculation
–$ 39.34
$176.48
$137.14
Estimate
–$ 40
$180
$140
The answer is not close to the estimate. It is not reasonable.
Additional Examples
Pre-AlgebraPre-Algebra
Add thefront-end digits.
You are buying some fruit. The bananas cost $1.32,
the apples cost $2.19, and the avocados cost $1.63. Use
front-end estimation to estimate the total cost of the fruit.
Rounding and EstimatingRounding and Estimating
Lesson 3-1
5.10=
The total cost is about $5.10.
+
Estimate byrounding.
.60
.20
.30
1.10
Additional Examples
Pre-AlgebraPre-Algebra
Estimate the total electricity charge: March: $81.75; April: $79.56; May: $80.89.
Rounding and EstimatingRounding and Estimating
Lesson 3-1
240=
The total electricity charge is about $240.00.
The values cluster around $80. 80
3 months
• 3
Additional Examples
Pre-AlgebraPre-Algebra
Objectives:1.To estimate products2. To estimate quotients
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
Pre-AlgebraPre-Algebra
Tips:On multiple choice questions, sometimes you can eliminate answers by estimating.
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
Pre-AlgebraPre-Algebra
Estimate 6.43 • 4.7.
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
Multiply.6 • 5 = 30
6.43 6 4.7 5 Round to the nearest integer.
6.43 • 4.7 30
Additional Examples
Pre-AlgebraPre-Algebra
Joshua bought 3 yd of fabric to make a flag. The fabric
cost $5.35/yd. The clerk said his total was $14.95 before tax.
Did the clerk make a mistake? Explain.
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
Multiply 5 times 3, the number ofyards of fabric.
5 • 3 = 15
5.35 5 Round to the nearest dollar.
The sales clerk made a mistake. Since 5.35 > 5, the actual cost should be more than the estimate. The clerk should have charged Joshua more than $15.00 before tax.
Additional Examples
Pre-AlgebraPre-Algebra
The cost to ship one yearbook is $3.12. The total cost
for a shipment was $62.40. Estimate how many books were in
the shipment.
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
3.12 3 Round the divisor.
The shipment is made up of about 20 books.
62.40 60 Round the dividend to a multiple of 3 that is close to 62.40.
60 ÷ 3 = 20 Divide.
Additional Examples
Pre-AlgebraPre-Algebra
Is 3.29 a reasonable quotient for 31.423 ÷ 5.94?
Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients
Lesson 3-2
Since 3.29 is not close to 5, it is not reasonable.
5.94 6 Round the divisor.
31.423 30 Round the dividend to a multiple of 6 that is close to 31.423.
30 ÷ 6 = 5 Divide.
Additional Examples
Pre-AlgebraPre-Algebra
Objectives:1.To find mean, median, mode, and range of a set of data.
2. To choose the best measure of central tendency.
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
Pre-AlgebraPre-Algebra
New Terms:1 Measures of Central Tendency – mean, median, mode of a collection of data.
2. Mean – is the sum of the data values divided by the number of data values, average.
3. Median – is the middle number when data values are written in order and there is an odd number of data values. For an even number of data values, the median is the mean of the two middle numbers.
4. Mode – is the data item that occurs most often. There can be one mode, more than one, or none.
5. Range – the difference between the greatest and least values in the data set.
6. Outlier – a data value that is much greater or less than the other data values.
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
Pre-AlgebraPre-Algebra
Six elementary students are participating in a one-week
Readathon to raise money for a good cause. Use the graph. Find
the (a) mean, (b) median, and (c) mode of the data if you leave
out Latana’s pages.
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
a. Mean:
40 + 45 + 48 + 50 + 505=
2335=
46.6=
The mean is 46.6.
sum of data valuesnumber of data values
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
b. Median: 40 45 48 50 50 Write the data in order.
The median is the middle number, or 48.
c. Mode: Find the data value that occurs most often.
The mode is 50.
Additional Examples
Pre-AlgebraPre-Algebra
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
a. $1.10 $1.25 $2.00 $2.10 $2.20 $3.50
No values are the same, so there is no mode.
b. 1 3 4 6 7 7 8 9 10 12 12 13
How many modes, if any, does each have? Name them.
c. tomato, tomato, grape, orange, cherry, cherry, melon, cherry, grape
There is one mode.
Both 7 and 12 appear more than the other data values.
Cherry appears most often.
Since they appear the same number of times, there are two modes.
Additional Examples
Pre-AlgebraPre-Algebra
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
a. Which data value is an outlier?
Use the data: 7%, 4%, 10%, 33%, 11%, 12%.
The data value 33% is an outlier. It is an outlier because it is 21% away from the closest data value.
b. How does the outlier affect the mean?
The outlier raises the mean by about 4 points.
12.8 – 8.8 = 4
Find the mean with the outlier.776 12.8
Find the mean without the outlier.445
8.8
Additional Examples
Pre-AlgebraPre-Algebra
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
a. the monthly amount of rain for a year
Which measure of central tendency best describes
each situation? Explain.
since the average monthly amount of rain for a year is not likely to have an outlier, mean is the appropriate measure.
Mean;
b. most popular color of shirt
Mode;
When the data have no outliers, use the mean.
When determining the most frequently chosen item, or when the data are not numerical, use the mode.
since the data are not numerical, the mode is the appropriate measure.
Additional Examples
Pre-AlgebraPre-Algebra
Mean, Median, and ModeMean, Median, and Mode
Lesson 3-3
c. times school buses arrive at school
(continued)
since one bus may have to travel much farther than other buses, the median is the appropriate measure.
Median;
When an outlier may significantly influence the mean, use the median.
Additional Examples
Pre-AlgebraPre-Algebra
Objectives: 1.To substitute into formulas2. To use the formula for the perimeter of a rectangle
Using FormulasUsing Formulas
Lesson 3-4
Pre-AlgebraPre-Algebra
New Terms:1. Formula – an equation that shows a relationship between quantities that are represented by variables.
2. Perimeter – the distance around a figure.
Using FormulasUsing Formulas
Lesson 3-4
Pre-AlgebraPre-Algebra
Using FormulasUsing Formulas
Lesson 3-4
Suppose you ride your bike 18 miles in 3 hours.
Use the formula d = r t to find your average speed.
d = r t Write the formula.
18 = (r )(3) Substitute 18 for d and 3 for t.
Divide each side by 3.183 =
3r
3
Simplify.6 = r
Your average speed is 6 mi/h.
Additional Examples
Pre-AlgebraPre-Algebra
Use the formula F = + 37, where n is the number
of chirps a cricket makes in one minute, and F is the
temperature in degrees Fahrenheit. Estimate the temperature
when a cricket chirps 76 times in a minute.
Using FormulasUsing Formulas
Lesson 3-4
n4
The temperature is about 56°F.
F = + 37n4 Write the formula.
F = + 37764 Replace n with 76.
F = 19 + 37 Divide.
F = 56 Add.
Additional Examples
Pre-AlgebraPre-Algebra
Find the perimeter of a rectangular tabletop with a
length of 14.5 in. and width of 8.5 in. Use the formula for the
perimeter of a rectangle, P = 2 + 2w.
Using FormulasUsing Formulas
Lesson 3-4
The perimeter of the tabletop is 46 in.
P = 2 + 2w Write the formula.
P = 46 Add.
P = 2(14.5) + 2(8.5) Replace with 14.5 and w with 8.5.
P = 29 + 17 Multiply.
Additional Examples
Pre-AlgebraPre-Algebra
Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals
Lesson 3-5
Objectives:
1. To solve one-step decimal equations involving addition
2. To solve one-step decimal equations involving subtraction
Pre-AlgebraPre-Algebra
Solve 6.8 + p = –9.7.
Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals
Lesson 3-5
6.8 + p = –9.7
6.8 – 6.8 + p = –9.7 – 6.8 Subtract 6.8 from each side.
p = –16.5 Simplify.
Check: 6.8 + p = –9.7
6.8 + (–16.5) –9.7 Replace p with –16.5.
–9.7 = –9.7
Additional Examples
Pre-AlgebraPre-Algebra
Ping has a board that is 14.5 ft long. She saws off a
piece that is 8.75 ft long. Use the diagram below to find the
length of the piece that is left.
Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals
Lesson 3-5
The length of the piece that is left is 5.75 ft.
x = 5.75 Simplify.
x + 8.75 – 8.75 = 14.5 – 8.75 Subtract 8.75 from each side.
x + 8.75 = 14.5
Additional Examples
Pre-AlgebraPre-Algebra
Solve –23.34 = q – 16.99.
Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals
Lesson 3-5
–23.34 = q – 16.99
–23.34 + 16.99 = q – 16.99 + 16.99 Add 16.99 to each side.
–6.35 = q Simplify.
Additional Examples
Pre-AlgebraPre-Algebra
Alejandro wrote a check for $49.98. His new account
balance is $169.45. What was his previous balance?
Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals
Lesson 3-5
Alejandro had $219.43 in his account before he wrote the check.
Simplify.p = 219.43
Add 49.98 to each side.p – 49.98 + 49.98 = 169.45 + 49.98
p – 49.98 = 169.45
Equation p – 49.98 = 169.45
Let p = previous balance.
Words previous balance minus check is new balance
Additional Examples
Pre-AlgebraPre-Algebra
Objectives:
1. To solve one-step decimal equations involving multiplication
2. To solve one-step decimal equations involving division
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
Pre-AlgebraPre-Algebra
Solve –6.4 = 0.8b.
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
–6.4 = 0.8b
–6.40.8
0.8b0.8= Divide each side by 0.8.
–8 = b Simplify.
Check: –6.4 = 0.8b
–6.4 0.8(–8) Replace b with –8.
–6.4 = –6.4
Additional Examples
Pre-AlgebraPre-Algebra
Every day the school cafeteria uses about 85.8
gallons of milk. About how many days will it take for the
cafeteria to use the 250 gallons in the refrigerator?
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
Let x = number of days.
numberof days
Words times is 250 gallonsdaily milk
consumption
Equation • =85.8 x 250
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
85.8x = 250
The school will take about 3 days to use 250 gallons of milk.
x = 2.914 . . . Simplify.
x 3 Round to the nearest whole number.
85.8x85.8 =
25085.8 Divide each side by 85.8.
Additional Examples
Pre-AlgebraPre-Algebra
Solve –37.5 = .
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
c–1.2
–37.5 =c
–1.2
Multiply each side by –1.2.–37.5(–1.2) = (–1.2)c
–1.2
45 = c Simplify.
Additional Examples
Pre-AlgebraPre-Algebra
A little league player was at bat 15 times and had
a batting average of 0.133 rounded to the nearest thousandth.
The batting average formula is batting average (a) = .
Use the formula to find the number of hits made.
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
hits (h)times at bat (n)
a =hn
0.133 =h
15 Replace a with 0.133 and n with 15.
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals
Lesson 3-6
0.133(15) = (15)h
15 Multiply each side by 15.
Simplify.1.995 = h
2 h Since h (hits) represents an integer, round to the nearest integer.
The little league player made 2 hits.
Additional Examples
Pre-AlgebraPre-Algebra
Using the Metric SystemUsing the Metric System
Lesson 3-7
Objectives:1.To identify appropriate metric measures2. To convert metric units
Pre-AlgebraPre-Algebra
Choose an appropriate metric unit. Explain your choice.
Using the Metric SystemUsing the Metric System
Lesson 3-7
a. the width of this textbook
b. the mass of a pair of glasses
c. the capacity of a thimble
Milliliter; a thimble will hold only a small amount of water.
Gram; glasses have about the same mass as many paperclips, but less than this textbook.
Centimeter; the width of a textbook is about two hands, or ten thumb widths, wide.
Additional Examples
Pre-AlgebraPre-Algebra
Choose a reasonable estimate. Explain your choice.
Using the Metric SystemUsing the Metric System
Lesson 3-7
a. capacity of a drinking glass: 500 L or 500 mL
b. length of a hair clip: 5 m or 5 cm
c. mass of a pair of hiking boots: 1 kg or 1 g
1 kg; the mass is about one half the mass of your math book.
5 cm; the length of a hair clip would be about 5 widths of a thumbnail.
500 mL; a drinking glass holds less than a quart of milk.
Additional Examples
Pre-AlgebraPre-Algebra
Complete each statement.
Using the Metric SystemUsing the Metric System
Lesson 3-7
a. 7,603 mL = L
7,603 ÷ 1,000 = 7.603 To convert from milliliters to liters, divide by 1,000.
7,603 mL = 7.603 L
4.57 m = 457 cm
To convert meters to centimeters, multiply by 100.
4.57 100 = 457 cm
b. 4.57 m = cm
Additional Examples
Pre-AlgebraPre-Algebra
A blue whale caught in 1931 was about 2,900 cm
long. What was its length in meters?
Using the Metric SystemUsing the Metric System
Lesson 3-7
Wordscentimeters per meter
length in centimeters ÷
length in meters=
2,900Equation 100 29÷ =
The whale was about 29 m long.
Additional Examples
Pre-AlgebraPre-Algebra
Objectives:1.To solve complex problems by first solving simpler cases
Problem Solving Strategy: Act It OutProblem Solving Strategy: Act It Out
Lesson 3-8
Pre-AlgebraPre-Algebra
Marta gives her sister one penny on the first day of October,
two pennies on the second day, and four pennies on the third day. She
continues to double the number of pennies each day. On what date will
Marta give her sister $10.24 in pennies?
Problem Solving Strategy: Act It OutProblem Solving Strategy: Act It Out
Lesson 3-8
1 2 2 • 2 = 4 4 • 2 = 8 8 • 2 = 1616 • 2 = 32
Number ofpennies
Days after the first
012345
Amount
$0.01$0.02$0.04$0.08 $0.16$0.32
Additional Examples
Pre-AlgebraPre-Algebra
(continued)
Problem Solving Strategy: Act It OutProblem Solving Strategy: Act It Out
Lesson 3-8
2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 = 1024
You can tell from the pattern in the chart that you just need to count the number of 2’s multiplied until you reach 1,024, which is $10.24 in pennies.
Marta will give her sister $10.24 in pennies on October 11.
10 twos = 10 days after the first penny is given
Additional Examples