Chapter 1, Problem 2CA

a.

To determine

To verify: The data shows a linear relationship, then write an equation of a line of fit.

Answer to Problem 2CA

The line of fit is $y=1462x+36208$ .

Explanation of Solution

Given Information: Table containing data-values of years and tuition.

Calculation:

  $\begin{array}{l}{m}_1=\frac{37620-36208}{1}=1412\\ \\ m_2=\frac{39088-37620}{2-1}=1468\\ \\ m_3=\frac{40594-39088}{3-2}=1506\end{array}$

As, slopes of different consecutive points do not vary much, this data can be formulated into linear equation of line of best fit.

But for the sake of simplicity, we will find line of best fit.

  $y=mx+b$

where $m=\frac{y_2-y_1}{x_2-x_1}$ ,

  $\begin{array}{l}m=\frac{40594-36208}{3}\\ \\ m=1462\end{array}$

When $x=0$ , $y=36208$ ,

  $\Rightarrow y-\mathrm{int}ercept=36208$

Substituting the values derived so far in the slope intercept form,

  $y=1462x+36208$

Thus, the line of fit is $y=1462x+36208$ .

b.

To determine

To interpret: The slope and $y-\mathrm{int}ercept$ in this situation.

Answer to Problem 2CA

Tuition fees is increasing by $\$1462$ every year, in $2010$ fees was $\$36208$ .

Explanation of Solution

Given Information: Equation from previous part.

Value of slope is implying that the tuition fees is increasing by $\$1462$ every year. Value of $y-\mathrm{int}ercept$ implies that in $2010$ fees was $\$36208$ .

c.

To determine

To predict: The cost of tuition in $2015$ .

Answer to Problem 2CA

Tuition fees will be $\$43518$ .

Explanation of Solution

Given Information: Equation from one of the previous parts.

Calculation:

  $y=1462x+36208$

Putting $x=5$ in above mentioned equation,

  $\begin{array}{l}y=1462(5)+36208\\ \\ =7310+36208\\ \\ =43518\end{array}$

Thus, tuition fees will be $\$43518$ .