Chapter 1.5, Problem 47E

Solution

a.

To Determine: A linear equation using the points $\left(52,70\right)$ and $\left(88,97\right)$ that can be used to convert raw scores to scaled grades.

The linear equation $y=31+\frac{3}{4}x$ can be used to convert raw scores to scaled grades. Here, $y$ represents the scaled grades and $x$ represents the raw scores.

Given:

The points $\left(52,70\right)$ and $\left(88,97\right)$ .Where the $x$ -coordinates represent the raw scores and the $y$ -coordinates represent the scaled grades.

Concepts Used:

The linear equation passing through the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Calculations:

The linear equation passing through the two points $\left(52,70\right)$ and $\left(88,97\right)$ is given by $\frac{y-70}{x-52}=\frac{97-70}{88-52}$ . Here, $y$ represents the scaled grades and $x$ represents the raw scores.

  $\begin{array}{c}\frac{y-70}{x-52}=\frac{97-70}{88-52}\\ \frac{y-70}{x-52}=\frac{27}{26}\\ \frac{y-70}{x-52}=\frac{3}{4}\\ y-70=\frac{3}{4}\left(x-52\right)\\ y=70+\frac{3}{4}x-\frac{3}{4}\times 52\\ y=70+\frac{3}{4}x-39\\ y=31+\frac{3}{4}x\end{array}$

Conclusion:

The linear equation $y=31+\frac{3}{4}x$ can be used to convert raw scores to scaled grades. Here, $y$ represents the scaled grades and $x$ represents the raw scores.

b.

To Determine: The inverse of the function defined by the linear equation determined in the first part.

To State: What the inverse function does.

The linear function $x=\frac{4}{3}y-\frac{124}{3}$ is the inverse $f^{-1}\left(y\right)$ of the linear function $f\left(x\right)=y=\frac{3}{4}x+31$ that converts raw scores to scaled grades.

The inverse function does the reverse operation, namely, it converts the scaled grades to the raw scores. Here, $y$ represents the scaled grades and $x$ represents the raw scores.

Given:

The points $\left(52,70\right)$ and $\left(88,97\right)$ .Where the $x$ -coordinates represent the raw scores and the $y$ -coordinates represent the scaled grades.

Known from previous part:

The linear equation $y=31+\frac{3}{4}x$ can be used to convert raw scores to scaled grades. Here, $y$ represents the scaled grades and $x$ represents the raw scores.

Concepts Used:

The inverse of a linear function $f\left(x\right)=y=mx+c$ is $f^{-1}\left(y\right)=x=\frac{y-c}{m}$ . Here, $m$ must be a nonzero real number.

Calculations:

The inverse of a linear function $f\left(x\right)=y=\frac{3}{4}x+31$ is $f^{-1}\left(y\right)=x=\frac{y-31}{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}$ .

  $\begin{array}{l}x=\frac{y-31}{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}\\ x=\frac{4}{3}\left(y-31\right)\\ x=\frac{4}{3}y-\frac{4}{3}\times 31\\ x=\frac{4}{3}y-\frac{124}{3}\end{array}$

Conclusion:

The linear function $x=\frac{4}{3}y-\frac{124}{3}$ is the inverse $f^{-1}\left(y\right)$ of the linear function $f\left(x\right)=y=\frac{3}{4}x+31$ that converts raw scores to scaled grades.

The inverse function does the reverse operation, namely, it converts the scaled grades to the raw scores. Here, $y$ represents the scaled grades and $x$ represents the raw scores.