Chapter 1.10, Problem 75E

a.

To determine

Find a linear equation that relates $C$ and $d$ .

Answer to Problem 75E

  $C=\frac{1}{4}d+260$

Explanation of Solution

Calculation:

The cost of driving $480\text{ mi}$ is $\$380$ for May.

The cost of driving $800\text{ mi}$ is $\$460$ for June.

The cost and distance are considered coordinates. The slope of the line is

  $\frac{y_2-y_1}{x_2-x_1}=\frac{460-380}{800-480}=\frac{1}{4}$

So, the line joining the points is

  $\begin{array}{l}C-380=\frac{1}{4}d-120\\ C=\frac{1}{4}d+260\end{array}$

Where $C$ is the cost of journey and $d$ is the distance travelled.

Hence, the linear equation that relates $C$ and $d$ is $C=\frac{1}{4}d+260$

b.

To determine

Use part $\left(a\right)$ to predict the cost of driving $\text{1500 mi}$ per month.

Answer to Problem 75E

  $\$635$

Explanation of Solution

Calculation:

When $d\text{=1500 mi}$ per month ,

  $\begin{array}{l}C=\frac{1}{4}\times 1500+260\\ C=\$635\end{array}$

Hence, the cost of driving $\text{1500 mi}$ per month is $\$635$

c.

To determine

Draw the graph of the linear equation. What does the slope of the line represent?

Answer to Problem 75E

  

The slope represents cost per mile.

Explanation of Solution

Calculation:

  

Note that the cost is represented on $y-axis$ and the distance travelled is represented on $x-axis$ .

Hence, the slope represents cost per mile.

d.

To determine

What does $y-\text{intercept}$ of the graph represents?

Answer to Problem 75E

The $y-\text{intercept}$ represents the monthly fixed cost.

Explanation of Solution

Calculation:

When $d=0$ ,

  $C=\$260$

which represents the monthly rental spent toward the car.

Hence, the $y-\text{intercept}$ represents the monthly fixed cost.

e.

To determine

Why is a linear relationship a suitable model for this situation?

Answer to Problem 75E

A linear fuction gives a suitable model because we would expect the cost of driving to be more or less proportional to the number of miles driven.

Explanation of Solution

Calculation:

A linear fuction gives a suitable model because we would expect the cost of driving to be more or less proportional to the number of miles driven.

Hence, a linear relationship is a suitable model for this situation.