Chapter 8 - Review Exercises - Page 951: 56

The coefficients are, $ a=\frac{5}{8},b=-50,c=1150\ $ and $ average=212.5$.

Consider that the given data set can be modeled by $ y=a{{x}^{2}}+bx+c $ Therefore $\begin{align} & a{{\left( 20 \right)}^{2}}+b\left( 20 \right)+c=400 \\ & a{{\left( 40 \right)}^{2}}+b\left( 40 \right)+c=150 \\ & a{{\left( 60 \right)}^{2}}+b\left( 60 \right)+c=400 \end{align}$ That is $\begin{align} & 400a+20b+c=400 \\ & 1600a+40b+c=150 \\ & 3600a+60b+c=400 \end{align}$ Therefore, the system of equations can be written in matrix form as below: $ AX=b $ Where $ A=\left[ \begin{array}{*{35}{r}} 400 & 20 & 1 \\ 1600 & 40 & 1 \\ 3600 & 60 & 1 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 400 \\ 150 \\ 400 \\ \end{array} \right];X=\left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]$ Therefore, using Cramer’s rule, the solution of the system of equations is given by: $ a=\frac{\left| {{A}_{a}} \right|}{\left| A \right|},b=\frac{\left| {{A}_{b}} \right|}{\left| A \right|},c=\frac{\left| {{A}_{c}} \right|}{\left| A \right|}$ Where ${{A}_{x}}=\left[ \begin{array}{*{35}{r}} 400 & 20 & 1 \\ 150 & 40 & 1 \\ 400 & 60 & 1 \\ \end{array} \right],{{A}_{y}}=\left[ \begin{array}{*{35}{r}} 400 & 400 & 1 \\ 1600 & 150 & 1 \\ 3600 & 400 & 1 \\ \end{array} \right],{{A}_{z}}=\left[ \begin{array}{*{35}{r}} 400 & 20 & 400 \\ 1600 & 40 & 150 \\ 3600 & 60 & 400 \\ \end{array} \right]$ Consider the determinant of the matrix $\begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 400 & 20 & 1 \\ 1600 & 40 & 1 \\ 3600 & 60 & 1 \\ \end{array} \right| \\ & =\left| \begin{array}{*{35}{r}} 400 & 20 & 1 \\ 1200 & 20 & 0 \\ 3200 & 40 & 0 \\ \end{array} \right| \\ & =1200\times 40-3200\times 20 \\ & =-16000 \end{align}$ And $\begin{align} & \left| {{A}_{x}} \right|=\left| \begin{array}{*{35}{r}} 400 & 20 & 1 \\ 150 & 40 & 1 \\ 400 & 60 & 1 \\ \end{array} \right| \\ & =\left| \begin{array}{*{35}{r}} 400 & 20 & 1 \\ -250 & 20 & 0 \\ 0 & 40 & 0 \\ \end{array} \right| \\ & =-250\times 40-0\times 20 \\ & =-10000 \end{align}$ And $\begin{align} & \left| {{A}_{y}} \right|=\left| \begin{array}{*{35}{r}} 400 & 400 & 1 \\ 1600 & 150 & 1 \\ 3600 & 400 & 1 \\ \end{array} \right| \\ & =\left| \begin{array}{*{35}{r}} 400 & 400 & 1 \\ 1200 & -250 & 0 \\ 3200 & 0 & 0 \\ \end{array} \right| \\ & =0+250\times 3200 \\ & =800000 \end{align}$ And $\begin{align} & \left| {{A}_{z}} \right|=\left| \begin{array}{*{35}{r}} 400 & 20 & 400 \\ 1600 & 40 & 150 \\ 3600 & 60 & 400 \\ \end{array} \right| \\ & =\left| \begin{array}{*{35}{r}} 400 & 20 & 400 \\ 0 & -40 & -1450 \\ 0 & -120 & -3200 \\ \end{array} \right| \\ & =400\left( 40\times 3200-120\times 1450 \right) \\ & =-18400000\text{ } \end{align}$ Therefore, the solution of the system of equations is given by $\begin{align} & a=\frac{-10000}{-16000}=\frac{5}{8} \\ & b=\frac{800000}{-16000}=-50 \\ & c=\frac{-18400}{-16}=1150 \end{align}$ Therefore, the fitted model $ y=\frac{5}{8}{{x}^{2}}-50x+1150$ Consider $\begin{align} & y\left( 30 \right)=\frac{5}{8}\times {{30}^{2}}-50\times 30+1150 \\ & =212.5 \\ & y\left( 50 \right)=\frac{5}{8}\times {{50}^{2}}-50\times 50+1150 \\ & =212.5 \end{align}$ Therefore, the average accident rate per day for a driver of age of 30 and 50 is 212.5. The predicted model is $ a=\frac{5}{8},b=-50,c=1150$