Scatter Plots

SCATTER PLOTS

The scatter plot shows education and income data.

a. Describe the person represented by the point with coordinates (10, 30).This person has 10 years of education and earns $30,000 each year.

b. How many people have exactly 14 years of education? What are their incomes?The points (14, 50), (14, 80), and (14, 90) have education coordinate 14.The three people they represent earn $50,000, $80,000, and $90,000, respectively.

SCATTER PLOTSUse the table to make a scatter plot of the elevation and precipitation data.

Atlanta, GABoston, MAChicago, ILHonolulu, HIMiami, FLPhoenix, AZPortland, MESan Diego, CAWichita, KS

Elevation Above Sea Level (ft)

1,05020

5961811

1,0727540

1,305

City Mean AnnualPrecipitation (in.)

5142362256

8441029

SCATTER PLOTSUse the scatter plot below. Is there a positive correlation, a negative correlation, or no correlation between temperatures and amounts of precipitation? Explain.

The values show no relationship.There is no correlation.

PROBLEM SOLVING STRATEGY: SOLVE BY GRAPHING

Use the data in the table below. Suppose this year there are 16 wolves on the island. Predict how many moose are on the island.

Isle Royale Populations

Wolf Moose

142324222016

700900811

1,0621,0251,380

Year

198219831984198519861987

Wolf Moose

121115121213

1,6531,3971,2161,3131,6001,880

Year

198819891990199119921993

Wolf Moose

151622241425

1,8002,4001,200

500700750

Year

199419951996199719981999

Step 1: Make a scatter plot by graphing the (wolf, moose) ordered pairs. Use the x-axis for wolves and the y-axis for moose.

Step 2: Sketch a trend line. The line should be as close as possible to each data point. There should be about as many points above the trend line as below it.


(continued)

Look up to find the point on the trend line that corresponds to 16 wolves.

Then look across to the value on the vertical axis, which is about 1,300.


(continued)

Step 3: To predict the number of moose when there are 16 wolves, find 16 along the horizontal axis.

There are about 1,300 moose on the island.

Graph each inequality on a coordinate plane.

GRAPHING LINEAR INEQUALITIES

a. y > 2x + 1

Step 1: Graph the boundary line.Points on the boundary line do not make y > 2x + 1 true. Use a dashed line.

(continued)


Step 2: Test a point not on the boundary line.Test (0, 0) in the inequality.

Since the inequality is false for (0, 0), shade the region that does not contain (0, 0).

y > 2x + 10 2(0) + 1 Substitute.0 0 + 1

0 > 1 false

>

>

(continued)


b. y 3x – 2<–

Step 1: Graph the boundary line.Points on the boundary line make y 3x – 2 true. Use a solid line.<–

(continued)


Step 2: Test a point not on the boundary line.Test (3, 0) in the inequality.

Since the inequality is true for (3, 0), shade the region containing (3, 0).

y 3x – 20 3(3) – 2 Substitute.0 9 – 20 7 true

<–

<–

<–?

<–?

Cashews cost $2/lb. Pecans cost $4/lb. You plan to spend no more than $20. How many pounds of each can you buy?


Step 1: Write an inequality.

Words cost of cashews plus cost of

pecansis at most

twenty dollars

Let = number of pounds of pecans.x

+Inequality 2y 4x <– 20

Let = number of pounds of cashews.y

(continued)


Step 3: Graph y = –2x + 10 in Quadrant I since weight is not negative.

Step 2: Write the equation of the boundary line in slope-intercept form.

2y + 4x 20y –2x + 10

<–

y = –2x + 10

<–

(continued)


Step 4: Test (1, 1).y –2x + 101 –2(1) + 101 8The inequality is true. (1, 1) is a solution.

<–<–?

<–

Step 5: Shade the region containing (1, 1).

The graph shows the possible solutions. For example, you could buy 1 pound of pecans and 5 pounds of cashews.

Step 1: Graph y x + 1 on a coordinate plane. Shade in red.

>–

Solve the system y x + 1 and y < 2x + 3 by graphing.

GRAPHING LINEAR INEQUALITIES>–

Step 2: Graph y < 2x + 3 on the same coordinate plane. Shade in blue.

(continued)


The solutions are the coordinates of all the points in the region that is shaded in both colors.

Check: See whether the solution (2, 5) makes both of the inequalities true.y x + 1 5 2 + 1 Replace x with 2 and y with 5.5 3y The solution checks.

y < 2x + 3 5 2(2) + 3 Replace x with 2 and y with 5.5 < 7y The solution checks.

>–>–?

>–

<?

Solve the system y = x – 7 and y = 4x + 2 by graphing.

SOLVING SYSTEMS OF LINEAR EQUATIONS

Step 1: Graph each line.Step 2: Find the point of intersection.

The lines intersect at one point, (–3, –10). The solution is (–3, –10).

y = x – 7

Check: See whether (–3, –10) makes both equations true.

–10 –3 – 7–10 = –10

y = 4x + 2–10 4(–3) + 2–10 = –10

Replace x with – 3and y with –10.

The solution checks.

Solve each system of equations by graphing.


a. 27x + 9y = 36; y = 4 – 3x b. 8 = 4x + 2y; 2x + y = 5

The lines are the same line.There are infinitely many solutions.

The lines are parallel.They do not intersect.There is no solution.

Find two numbers with a sum of 10 and a difference of 2.


Step 1: Write equations.Let x = the greater number.Let y = the lesser number.

Equation 1 Sum is 10.x + y = 10Equation 2 Difference is 2.

x – y = 2Step 2: Graph the equations.

The lines intersect at (6, 4).The numbers are 6 and 4.

(continued)


Check: Since the sum of 6 and 4 is 10 and the difference of 6 and 4 is 2, the answer is correct.

Scatter Plots

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