INTRODUCTION

In chapter 2 you will learn how to solve equations and inequalities involving a single variable. In this

chapter we are concerned with equation or inequalities form a system. In section 6.1 and 6.2 we solve

system of linear equations, whereas in section 6.3 we solve system involving quadratic equations.

Systems of linear inequalities are created in section 6.4, and in the same section there is an introduction to

linear programming.

6.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES

Many applications of mathematics lead to more than one equation in several variables. The resulting

equations are called a system of equations. The solution set of a system of equations consists of all

solutions that are common to the equation in the system.

ax + by = c

We proved that the graph of an equation of the form is a line and that all ordered pairs (x , y) satisfying

the equation are coordinates of points in the line. A system of two linear equations in two variables x and

y can be written as.

{a1x + b1y = c1}

{a2x + b2y = c2}

Where a1,b1, c1, a2, b2,and c2 are real numbers. The left brace is used to indicate that the two equations

form a system. If an ordered pair (x , y) is to satisfy the system of two linear equations, the corresponding

points (x , y) must lie on the two lines that are the graph of the equations.

ILLUSTRATION 1

A particular system of two linear equations is.

{2𝑥 + 𝑦 = 3

5𝑥 + 3𝑦 = 10

The solution set of each of the equations in the system is an infinite set of ordered pairs of real numbers,

and the graphs of these sets are lines. Recall that to draw a sketch of a line we need to find two points on

the line; usually we plot the points where the line intersects the coordinate axis. On the line 2x + y = 3 we

have the points (3/2 , 0) and (0,3) while on the line 5x + 3y = 10 we have the points (2,0) and(0, 10/3).

Shows on the same coordinate system sketches of two lines. It is apparent that two lines intersect at

exactly one point. This point, (-1 , 5) can be verified by substituting into the equations as follows:

2(-1) + 5 = 3

5(-1) + 3(5) = 10

The only ordered pair that is common to the solution sets of the two equations is (-1,5). Hence the

solution set of the system is {(-1,5)}.

ILLUSTRATION 2

Consider the system

{6𝑥 − 3𝑦 = 52𝑥 − 𝑦 = 4

The lines having these equations appear to be parallel. It can easily to be proved that the lines are indeed

parallel by writing each of the equation, we have

6x - 3y = 5 2x – y = 4

-3y = -6x + 5 -y = -2x + 4

y = 2x – 5/3 y = 2x – 4

Example 1

Use the substitution method to find the solution set of the system.

Illustration 1: {2𝑥 + 𝑦 = 3

5𝑥 + 3𝑦 = 10

Solution:

We solve the first equation for y and get the equivalent system

{𝑦 = 3 − 2𝑥

5𝑥 + 3𝑦 = 10

We replace y in the second equation by its equal, 3 – 2 x, from the first equation. We then have the

equivalent system

{𝑦 = 3 − 2𝑥

5𝑥 + 3(3 − 2𝑥 ) = 10

Simplifying the second equation, we have

{𝑦 = 3 − 2𝑥

−𝑥 + 9 = 10

Solving the second equation for x, we get

{𝑦 = 3 − 2𝑥

𝑥 = −1

Finally, we substitute the value of x from the second equation into the first equation and we have

{𝑦 = 5

𝑥 = −1

This system is equivalent to the given one. Hence the solution set is ( -1 , 5)

Example 2

Use the elimination method to find the solution set of the system of equations in Example 1.

{2𝑥 + 𝑦 = 3

5𝑥 + 3𝑦 = 10

Remember that our goal is to eliminate one of the variables. Observe that the coefficient

of y is 1 in the first equation and 3 in the second equation. To obtain an equation not involving y, we

therefore replace the second equation by the sum of the second equation and -3 times the first. We begin

by multiplying the first equation by -3 and writing the equivalent system.

{ −6 − 3𝑦 = −95𝑥 + 3𝑦 = 10

Adding the equations given the following computations:

−6−3𝑦 = −95𝑥 +3𝑦 =10

−𝑥 =1

With this equation and the first equations in the given system, we can write the following equivalent

system

{2𝑥 + 𝑦 = 3 −𝑥 = 1

If we now multiply both sides of the second equation by -1, we have the equivalent system

{2𝑥 + 𝑦 = 3

𝑥 = −1

We next substitute -1 for x in the first equation to obtain

{2(−1) + 𝑦 = 3 𝑥 = −1

{𝑦 = 5

𝑥 = −1

Therefore the solution set (-1,5), which agrees with the result of example 1.

Exercises 6.1 Show the solutions

1.) {𝑥 − 𝑦 = 8

2𝑥 + 𝑦 = 1 2.) {

𝑦 = 8 + 2𝑥6𝑥 + 3𝑦 = 0

3.) {2𝑥 + 𝑦 = 6

8𝑥 = 6𝑦 + 9 4.) {

9𝑥 − 3𝑦 = 7𝑦 = 3𝑥 − 5/2

5.) {𝑦 = 2𝑥 − 4

6𝑥 − 3𝑦 − 12 = 0 6.) {

2𝑥 − 3𝑦 = −15𝑥 − 4𝑦 = 8

7.) {4𝑥 − 2𝑦 − 7 = 0

𝑥 =1

2𝑦 + 5

8.) {3𝑥 − 𝑦 = 1

6𝑥 + 5𝑦 = 2

9.) {2𝑥 + 6𝑦 = −114𝑥 − 3𝑦 = −2

10.) {𝑦 = 3𝑥 − 5

6𝑥 − 2𝑦 = 10

6.2 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES

So far the linear (first degree) equations we have discussed have contained at most two variables. In this

section we introduce systems of linear equations in three variables.

Consider the equation

2𝑥 − 𝑦 + 4𝑧 = 10

For which the replacement set of each of the three variables x, y and z is the set R of real numbers. This

equation is linear in the three variables. A solution of a linear equation in the three variables x, y and z is

the ordered triple of real numbers (r,s,z) such that if x is replaced by r, y by s, and z by t, the resulting

statement is true. The set of all solutions is the solution set of the equation

Illustration 1

For the equation

2𝑥 − 𝑦 + 4𝑧 = 10

The ordered triple pair (3,4,2) is a solution because

2(3) – 4 + 4 (2) = 10

Some other ordered triples that satisfy this equation are (-1, 8, 5) , (2, -6, 0), (1,0,2), (5,0,0) , (0,-6,1) , (8,

2 1) ,and (7,2 - ½). It appears that the solution set is infinite.

The graph of an equation in three variables is a set of points represented by ordered triples of real

numbers. Such points appear in a three dimensional coordinate system, which we do not discuss. You

should, however, be aware that the graph of a linear equation in a three variables is a plane.

Suppose that we have the following system of linear equations in the variables x, y and z.

{

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑎2𝑥 + 𝑏2𝑦 + 𝑐2𝑧 = 𝑑2𝑎3𝑥 + 𝑏3𝑦 + 𝑐3𝑧 = 𝑑3

The solution set of this system is the intersection of the solution sets of the three equations. Because the

graph of each equation is a plane, the solution set can be interpreted geometrically as the intersection of

three planes. When this intersection consist and independent.

Algebraic methods for finding the solution set of a system of three linear equations in three variables are

analogous to those used to solve linear systems in two variables. The following examples shows the

substitution method.

Example Find the solution set of the system

{𝑥 − 𝑦 − 4𝑧 = 3

2𝑥 − 3𝑦 + 2𝑧 = 02𝑥 − 𝑦 + 2𝑧 = 2

Solution, we solve the first set of the system

{𝑥 = 𝑦 + 4𝑧 + 3

2𝑥 − 3𝑦 + 2𝑧 = 02𝑥 − 𝑦 + 2𝑧 = 2

We now substitute the value of x from the first equation into the other two equations , and we obtain the

equivalent system

{

𝑥 = 𝑦 + 4𝑧 + 3

2(𝑦 + 4𝑧 + 3) − 3𝑦 + 2𝑧 = 0

2(𝑦 + 4𝑧 + 3) − 𝑦 + 2𝑧 = 2

{𝑥 = 𝑦 + 4𝑧 + 3−𝑦 + 10𝑧 = −6𝑦 + 10𝑧 = −4

We next solve the second equation for y and get

𝑥 = 𝑦 + 4𝑧 + 3

𝑦 = 10𝑧 + 6

𝑦 + 10𝑧 = −4

Substituting the value of y from the second equation into the third gives the equivalent system.

{

𝑥 = 𝑦 + 4𝑧 + 3𝑦 = 10𝑧 + 6

(10𝑧 + 6 +) + 10 = −4

{𝑥 = 𝑦 + 4𝑧 + 3

𝑦 = 10𝑧 + 620𝑧 = −10

{

𝑥 = 𝑦 + 4𝑧 + 3𝑦 = 10𝑧 + 6

𝑧 = −1/2

Substituting the value of z from the third equation into the second equation, we obtain

{

𝑥 = 𝑦 + 4𝑧 + 3𝑦 = 1

𝑧 = −1/2

Substituting the values of y and z from the second and third equations into the first equation, we get

{𝑥 = 2𝑦 = 1

𝑧 = 1/2

The latter system is equivalent to the given system. Hence the solution set of the given system is (2,1

,1/2).The solution can be checked by substituting into each of the given equations. Doing this we have

{2 − 1 + 2 = 34 − 3 − 1 = 04 − 1 − 1 = 2

The equations of the given system are consistent and independent.

Exercise

1.) {4𝑥 + 3𝑦 + 𝑧 = 15

𝑥 − 𝑦 − 2𝑧 = 22𝑥 − 2𝑦 + 𝑧 = 4

2.) {

2𝑥 + 3𝑦 + 𝑧 = 85𝑥 + 2𝑦 + 3𝑧 = −13

𝑥 − 2𝑦 + 5𝑧 = 15

3.) {𝑥 − 𝑦 + 3𝑧 = 2

2𝑥 + 2𝑦 − 𝑧 = 55𝑥 + 2𝑧 = 7

4.) {3𝑥 + 2𝑦 − 𝑧 = 43𝑥 + 𝑦 + 3𝑧 = −2

6𝑥 − 3𝑦 − 2𝑧 = −6

5.) {2𝑥 − 3𝑦 − 5𝑧 = 4𝑥 + 7𝑦 + 6𝑧 = −77𝑥 + 2𝑦 − 9𝑧 = 6

6.) {

3𝑥 − 2𝑦 + 4𝑧 = 47𝑥 − 5𝑦 − 𝑧 = 9𝑥 + 9𝑦 − 9𝑧 = 1

7.) {3𝑥 − 5𝑦 + 2𝑧 = −2

2𝑥 + 3𝑧 = −34𝑦 − 3𝑧 = 8

8.) {𝑥 − 𝑦 = 23𝑦 + 𝑧 = 1𝑥 − 2𝑧 = 7

9.) {3𝑥 − 2𝑦 = 1

𝑧 − 𝑦 = 5𝑧 − 2𝑥 = 5

10.) {𝑥 − 𝑦 + 5𝑧 = 2

4𝑥 − 3𝑦 + 5𝑧 = 33𝑥 − 2𝑦 + 4𝑧 = 1

6.3 SYSTEMS INVOLVING QUADRATIC EQUATIONS

In sections 6.1 and 6.2 our discussions of systems of equations was confined to linear systems. However,

a number of applications lead to nonlinear systems as illustrated in exercises 25 through 36. The word

problems in these exercises use concepts presented previously, but the resulting systems involve at least

one quadratic equation. In this section we discussed methods of solving such systems of two equations in

two variables.

We consider first a system that contains a linear equation and quadratic equation. In this case the system

can be solved for one variable in terms of the other, and the resulting expression can be substituted into

the quadratic equation, as shown in the following example.

Example 1

Find the solution set of the system.

{𝑦2 = 4𝑥

𝑥 + 𝑦 = 3

Solution

We solve the second equation for x and obtain the equivalent system.

{𝑦2 = 4𝑥

𝑥 = 3 − 𝑦

Replacing x in the first equation by its equal from the second, we have the equivalent system

{𝑦2 = 4(3 − 𝑦)

𝑥 = 3 − 𝑦

{𝑦2 + 4𝑦 − 12 = 0

𝑥 = 3 − 𝑦

We now solve the first equation.

(𝑦 − 2)(𝑦 + 6) = 0

𝑦 − 2 = 0 𝑦 + 6 = 0

y= 2 𝑦 = −6

Because the first equation of system (II) is equivalent to the equations 𝑦 = 2 and 𝑦 = −6, system (II) is

equivalent to the systems

{𝑦 = 2

𝑥 = 3 − 𝑦 and {

𝑦 = −6𝑥 = 3 − 𝑦

In each of the latter two systems we have substitute into the second equation the value of y from the first ,

and we have

{𝑦 = 2𝑥 = 1

and {𝑦 = −6𝑥 = 9

These two systems are equivalent to system (I). Thus, the solution set of (I) is (1,2) , (9,-6).

Exercises

Find the solution set of the system.

1.) {𝑥2 + 𝑦2 = 25𝑥 − 𝑦 + 1 = 0

2.) {𝑥2 + 𝑦2 = 25𝑥 − 2𝑦 = −2

3.) {𝑥2 − 𝑦 = 1

𝑥2 − 2𝑦 = −1 4.) {

𝑥2 − 𝑦2 = 92𝑥 + 𝑦 = 6

5.) {𝑥2 − 𝑦2 = 9

𝑥 + 𝑦 = 5 6.) {

4𝑥2 + 𝑦2 = 252𝑥 + 𝑦 + 1 = 0

7.) {𝑥2 − 𝑦 − 4 = 0𝑥 − 𝑦 − 3 = 0

8.) {4𝑥2 + 𝑦 − 3 = 08𝑥 + 𝑦 − 7 = 0

9.) {𝑥2 − 2𝑦2 = 2

𝑥 + 2𝑦 = 2 10.) {

4𝑥2 + 𝑦2 = 17

𝑥2 + 𝑦 = 5

6.4 SYSTEMS OF LINEAR INEQUALITIES AND INTRODUCTION TO LINEAR

PROGRAMMING

Systems of linear inequalities are important in economics, business, statistics, science, engineering, and

other fields. With electronic computers performing most of the computation, large numbers of inequalities

with many unknowns are usually involved. In this section we briefly discuss how to solve system of

linear inequalities. We then give an introduction to linear programming, a related approach to decision

making problems.

Statement of the form

𝐴𝑥 + 𝐵𝑦 + 𝐶 > 0 𝐴𝑥 + 𝐵𝑦 + 𝐶 < 0

𝐴𝑥 + 𝐵𝑦 + 𝐶 ≥ 0 𝐴𝑥 + 𝐵𝑦 + 𝐶 ≤ 0

Where A,B and C are constants, A and B are not both zero, are inequalities of first degree in two

variables. By the graph of such an inequality, we mean the (x, y) in the rectangular Cartesian coordinate

system for which (x, y) is an ordered pair satisfying the inequality.

Every line in a plane divides the plane into two regions, one on each side of the line. Each of these

regions is called a half plane. The graphs of inequalities of the forms.

𝐴𝑥 + 𝐵𝑦 + 𝐶 > 0 and 𝐴𝑥 + 𝐵𝑦 + 𝐶 < 0

Are half planes. We shall show this for the particular inequalities

2𝑥 − 𝑦 − 4 > 0 𝑎𝑛𝑑 𝐴𝑥 + 𝐵𝑦 + 𝐶 < 0

Let L be the line having the equation 2𝑥 − 𝑦 − 4 = 0. If we solve this equation for y, we obtain 𝑦 =

2𝑥 − 4. If (x, y) is any point in the plane, exactly one of the following statements holds:

𝑦 = 2𝑥 − 4 𝑦 > 2𝑥 − 4 𝑦 < 2𝑥 − 4

Now, 𝑦 > 2𝑥 − 4 if and only if the point (x, y) is any point (x, 2x - 4) on line L; Furthermore, 𝑦 < 2𝑥 −

4 if and only if the point (x, y) is below the point (x, 2x - 4) on L; therefore the line L divides the plane

into two regions. One region is the half plane above L, which is the graph of inequality 𝑦 > 2𝑥 − 4, and

the other region is the half plane above L, which is the graph of the inequalities 𝑦 > 2𝑥 − 4, and the

region is the half plane below L, which is the graph of inequality 𝑦 < 2𝑥 − 4. A similar discussion holds

for any line L having an equation of the form 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 𝑤ℎ𝑒𝑟𝑒 𝐵 ≠ 0.

If B= 0, an equation of line L is 𝐴𝑥 + 𝐶 = 0, and L is a vertical line whose equation 𝑥 = 4. Then if

(x, y) is any point in the plane, exactly one of the following statements is true:

The point (x, y) is to the right of the point (4, y) if and only if x >4. Showing the graph of inequality x >4

as the half plane lying to the right of the line x = 4. Similarly, the graph of x < 4 if, and only if the point

(x, y) is to the left of the point (4, y). The discussion can be extended to any line having an equation of

the form Ax +𝐶 = 0.

By generalizing the above arguments to any line, we can prove this theorem.

THEOREM

(I) the graph of y> 𝑚𝑥 + 𝑏 is the half plane lying above the line y= 𝑚𝑥 + 𝑏.

(II) the graph of y< 𝑚𝑥 + 𝑏 is the half plane lying below the line y= 𝑚𝑥 + 𝑏.

(III) the graph of (y< 𝑚𝑥 + 𝑏) x> 𝑎 is the half plane lying to the right of line x= 𝑎.

(IV) The graph of x< 𝑎 is the half plane lying to the left of the line x= 𝑎.

Example 1

Draw a sketch of the graph of the inequality

2𝑥 − 4𝑦 + 5 > 0

Solution

The given inequality is equivalent to

−4𝑦 > −2𝑥 − 5

𝑦 >1

2 + 5/4

The graph of inequality is the half plane below the line having the equation 𝑦 = 1/2𝑥 + 5/4. A sketch of

this graph is the shaded half plane.

A closed half plain is a half plane together with the line bounding it is the graph of an inequality of the

form.

𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 𝑜𝑟 𝐴𝑥 + 𝐵𝑦 + 𝐶 ≤ 0

Illustration

The inequality

4𝑥 + 5𝑦 − 20 ≥ 0

Is equivalent to

5𝑦 ≥ −4𝑥 + 20

𝑦 ≥ −4/5𝑥 + 4

Therefore the graph of this inequality is the closed half plane consisting of the line 𝑦 = −3/5𝑥 + 4 and the

half plane above it. A sketch of the graph.

Two intersecting lines divide the points of the plane into four regions. Each of these regions is the

intersection of two half planes and is defined by a system of two linear inequalities.

Exercise

Draw the sketch of the graph of inequality

1. (a) x> 2; (𝑏)𝑥 ≤ −3 2. (𝑎)𝑦 ≥ −5; (𝑏)𝑦 < 6

3. (𝑎)𝑥 ≥ 4; (𝑏)𝑥 ≤ 7 4. (𝑎)𝑦 > −1; (𝑏)𝑦 ≤ 8

5. 𝑦 < 4𝑥 − 2 6. 𝑦 ≥ 2𝑥 − 3

7. 2𝑥 − 7 ≥ 0 8. 𝑦 + 8 < 0

9. 5𝑥 + 6 > 2𝑦 10. 2𝑦 − 8𝑥 + 5 > 0

CHAPTER 8

STATISTICS AND PROBABILITY

APPLICATION IN PROBABILITY

Probability is a way to measure the chances that something will occur in relation to the possible

alternatives. For example, the probability is not a guarantee. A couple might have six children and all are

boys, or they might have six children and all are girls.

Now you might think that a couple with six girls would not expect to have another girl if they decided to

have a seventh child. In fact, the probability that the seventh child is a girl is still ½ , since the gender of this

child is not affected by the gender of the previous six children.

What if you know that a family has seven children and six of them are girls? Is the probability that the

seventh child is a girl still1/2? There are eight possibilities: all seven children are girls, or either the first,

second, third, fourth, fifth, sixth, or seventh child is a boy. Since in only one of these cases the seventh child

is a girl, the probability is 1/8.

14-1 STATISTICS AND LINE PLOTS

Objectives After studying this lesson, you should be able to:

Interpret numerical data from a table, and

Display and interpret statistical data on a line plot.

Each day when you read news papers or magazines, watch television, or listen to the radio, you are

bombarded with numerical information about the national economy, sports, politics, and so on.

Interpreting this numerical information, or data, is important to your understanding of the world around

you. A branch of mathematics called statistics helps provide you with methods for collecting, organizing

and interpreting data.

Statistical data can be organized and

presented in numerous ways. One of the most

common ways is to use a table or chart. The

chart at the right shows the hourly wages

earned by the principal wage earner in ten

families.

Using tables or charts like the one at the right should enable you to more easily analyze the given data.

Example 1

Use the information in the chart above to answer each question.

a. What are the maximum and minimum hourly wages of the principal wage earner for the ten families.

The families.

b. What percent of the families have a principal wage earner that makes less than $ 10.00 per hour.

In some instances, statistical data can be presented on a number line. Numerical information displayed on a

number line is called a line plot. For example, the data in the table above can presented in a line plot.

For example 1, you know that the data in the chart range from $ 8.00 per hour to $ 20.25 per hour. In order

to represent each hourly wage on a number line, the scale used must includes these values . A “w” is use to

represent each hourly wage. When more than one “w” s has the same location on the number line, additional

“w” are placed one above the other. A line plot for the hourly wages is shown below.

Exercises

Practice

1. Use the line plot at the right to answer each question.

a. What was the highest score on the test?

b. What was the lowest score on the test?

c. How many students took the test?

d. How many students scored in the 40’s?

e. What score was received by the most students?

Family Hourly wage

A $ 8.00 B $ 10.50

C $ 20.25

D $ 9.40 E $11.00

F $ 13.75 G $ 8.50

H $ 10.50 I $ 9.00

J $ 11.00

20-25 3

25-30 10 30-35 8

35-40 2 40-45 7

45-50 5

50 1

14.2 STEM AND LEAF PLOTS

Objective : After studying this lesson, you should be able to:

Display and interpret data on a stem-leaf-plot.

Application: Mr.Juaez wants to study the distribution of the scores for a 100-point unit exam given in his first-

period Biology class. The scores of the 35 students in the class are listed below.

82 77 49 84 44 98 93 71 76 65 89 95 78 69 89 64

88 54 96 87 92 80 44 85 93 89 55 62 79 90 86 75

74 99 62

He can organize and display the scores in a compact way using a stem-leaf-plot.

In a stem-leaf-plot, the greatest common place value of the data is used to form the stems. The numbers in the

next common place-value position are then used to form the leaves. In the list above, the greatest place value is

tens. Thus, the number 82 would have stem 8 and leaf 2.

To make the stem-and-leaf-plot, first make

a vertical list of the stems. Since the test

scores range from 44-99, the stems range

from 4-9. Then, plot each number by

placing leaf 2 to the right of the stem 8. The

right.

A second stem-and-leaf-plot can be made

To arrange the leaves in numerical order from

Least to greatest as shown at the right. This

will make it easier for Mr. Juarez to analyze

the data.

Example 1

STEM LEAF

4 9 4 4

5 4 5

6 5 9 4 2 2 7 7 1 6 8 9 5 4

8 2 4 9 9 8 7 0 5 9 6 9 8 3 5 6 2 3 0 9

8/2 Represents a score of 82

STEM LEAF

4 4 4 9

5 4 5

6 2 2 4 5 9 7 1 4 5 6 7 8 9

8 0 2 4 5 6 7 8 9 9 9

9 0 2 3 3 5 6 8 9

Use the information in the stem-and-leaf-plots above to answer each question. a. What were the highest and lowest scores on the test? 99 and 44 b. Which test scores occurred most frequently? 89 (3times) c. In which 10-point interval did the most students score? 80-89 (10 students) d. How many students received a score of 70 or better? 25 students

Sometimes the data for a stem-and-leaf-plot are numbers that have more than two digits. Before plotting these

numbers, they may need to be rounded or truncated to determine each stem and leaf. Suppose you wanted to

plot 356 using the hundreds digit for the stem.

ROUNDED TRUNCATED

Rounded 356 to 360. Thus, you To truncate means to cut off, so

Would plot 356 using stem 3 truncate 356 as 350. Thus, you

And leaf 6. What would be the would plot 356 using stem 3 and leaf

Stem and leaf of 499? 5 and 0 5. What would be the stem and leaf of 499? 4 and 9

A back-to-back stem-and-leaf plot is sometimes used to compare two sets of data or rounded and truncated

values of the same set of data. In a back-to-back plot, the same stem is used for the leaves of both plot.

Exercises

Applications

1. Football The stem-and-leaf plot below gives the number of catches of the NFL’s leading pass receiver for

each season through 1990.

a. What was the greatest number of

catches during the season?

b. What was the least number of

catches during the season?

c. How many seasons are listed?

d. What number of catches occurred

most frequently?

e. How many times did the leading

pass receiver have at least 90

catches?

STEM LEAF

6 0 1 2 6 7 7 1 1 1 2 2 3 3 3 4 5 8

8 0 2 6 8 8 9 9 0 1 2 2 5

10 0 0 1 6 9/2 Represents 92 catches