# A multiple regression equation to predict a student's score in College Algebra (hat{y}) based on their high school GPA

A multiple regression equation to predict a student's score in College Algebra $\left(\stackrel{^}{y}\right)$ based on their high school GPA (${x}_{1}{x}_{1}$), their high school Algebra II grade (${x}_{2}{x}_{2}$), and their placement test score (${x}_{3}{x}_{3}$) is given by the equation below.
$\stackrel{^}{y}=-9+5{x}_{1}{x}_{1}+6{x}_{2}{x}_{2}+0.3{x}_{3}{x}_{3}$
a) According to this equation, what is the predicted value of the student's College Algebra score if their high school GPA was a 3.9, their high school Algebra II grade was a 2 and their placement test score was a 40? Round to 1 decimal place.
b) According to this equation, what does the student's placement test score need to be if their high school GPA was a 3.9, their high school Algebra II grade was a 2, and their predicted College Algebra score was a 67? Round to 1 decimal place.

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Step 1
Given,
A multiple regression equation to predict a student's score in College Algebra $\left(\stackrel{^}{y}\right)$ based on their high school GPA $\left({x}_{1}\right)$, their high school Algebra II grade $\left({x}_{2}\right)$, and their placement test score $\left({x}_{3}\right)$ is given by the equation below.
$\stackrel{^}{y}=-9+5{x}_{1}+6{x}_{2}+0.3{x}_{3}$
Step 2
a)
The predicted value of the student's College Algebra score if their high school GPA was a 3.9, their high school Algebra II grade was a 2 and their placement test score was a 40 is calculated as follows:
$\stackrel{^}{y}=-9+5{x}_{1}+6{x}_{2}+0.3{x}_{3}$
$=-9+5\left(3.9\right)+6\left(2\right)+0.3\left(40\right)$
$=-9+43.5=34.5$
The predicted value of the student's College Algebra score is 34.5.
b)
The student's placement test score need to be if their high school GPA was a 3.9, their high school Algebra II grade was a 2, and their predicted College Algebra score was a 67 is calculated as follows:
$\stackrel{^}{y}=-9+5{x}_{1}+6{x}_{2}+0.3{x}_{3}$
$67=-9+5\left(3.9\right)+6\left(2\right)+0.3{x}_{3}$
$67=22.5+0.3{x}_{3}$
$44.5=0.3{x}_{3}$
${x}_{3}=\frac{44.5}{0.3}=148.3$
The student's placement test score need to be 148.3

Jeffrey Jordon