Chapter 8.3, Problem 72E

a.

To determine

The system of the linear equations for the data.

Answer to Problem 72E

  $\begin{array}{l}64a+8b+c=2556\\ 81a+9b+c=2742\\ 100a+10b+c=3250\end{array}$

Explanation of Solution

Given information:

The table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

  $(a)$ The data can be modelled by the quadratic function $y=a{t}^2+bt+c$ . Create a system of linear equations for the data. Let $t$ represent the year, with $t=8$ corresponding to $2008$ .

Calculation:

From the given information,the table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

The data can be modelled by the quadratic function $y=a{t}^2+bt+c$ will create a system of linear equations for the data. Let $t$ represent the year, with $t=8$ corresponding to $2008$ .

  $\begin{array}{l}a{(}^8+b(8)+c=2556\\ 64a+8b+c=\mathrm{2556......................}(1)\\ a{(}^9+b(9)+c=2742\\ 81a+9b+c=\mathrm{2742......................}(2)\\ a{(}^{10}+b(10)+c=3250\\ 100a+10b+c=\mathrm{3250......................}(3)\end{array}$

The system of equations are:

  $\begin{array}{l}64a+8b+c=2556\\ 81a+9b+c=2742\\ 100a+10b+c=3250\end{array}$

b.

To determine

The least square regression parabola.

Answer to Problem 72E

  $y=161t^2-2551t+12660$ .

Explanation of Solution

Given information:

The table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

  $(b)$ Use the matrix capabilities of a graphing utility to find the inverse matrix to solve the system from part (a) and find the least squares regression parabola $y=a{t}^2+bt+c$ .

Calculation:

To find the least square regression line write the system in matrix form $AX=B$:

  $A=\left[\begin{array}{ccc}64& 8& 1\\ 81& 9& 1\\ 100& 10& 1\end{array}\right],X=\left[\begin{array}{c}a\\ b\\ c\end{array}\right],B=\left[\begin{array}{c}2556\\ 2742\\ 3250\end{array}\right]$

Find the inverse of $A$ .Form the matrix $\left[A|I_3\right]$ :

  $\left[\begin{array}{ccc}64& 8& 1\\ 81& 9& 1\\ 100& 10& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Identify the inverse using row operations by multiplying the first row by $\frac{1}{64}$ :

  $\left[\begin{array}{ccc}1& 0.125& \frac{1}{64}\\ 81& 9& 1\\ 100& 10& 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{64}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Multiply the first row by $81$ and after that subtract the first row from the second row, and multiply the first row by $100$ and after that subtract the row from the third row:

  $\left[\begin{array}{ccc}1& 0.125& \frac{1}{64}\\ 0& -1.125& -\frac{17}{64}\\ 0& -2.5& \frac{-9}{16}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{64}& 0& 0\\ \frac{-81}{64}& 1& 0\\ \frac{-25}{16}& 0& 1\end{array}\right]$

Multiply the second row by $\frac{-8}{9}$ and after that multiply the second row by $2.5$ and add with third row:

  $\left[\begin{array}{ccc}1& 0.125& \frac{1}{64}\\ 0& 1& \frac{17}{72}\\ 0& 0& \frac{1}{36}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{64}& 0& 0\\ 1.125& \frac{-8}{9}& 0\\ 1.25& \frac{-20}{9}& 1\end{array}\right]$

Multiply the third row with $36$ and after that multiply the third row with $\frac{17}{72}$ and subtract from the second row:

  $\left[\begin{array}{ccc}1& 0.125& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\frac{-11}{16}& 1.25& \frac{-9}{16}\\ -9.5& 18& -8.5\\ 45& -80& 36\end{array}\right]$

Multiply the second row by $0.125$ and subtract from the first row:

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}0.5& -1& 0.5\\ -9.5& 18& -8.5\\ 45& -80& 36\end{array}\right]$

The solution to the system is $X=A^{-1}B$ :

  $\begin{array}{l}\left[\begin{array}{ccc}0.5& -1& 0.5\\ -9.5& 18& -8.5\\ 45& -80& 36\end{array}\right]\left[\begin{array}{l}2556\\ 2742\\ 3250\end{array}\right]\\ X=\left[\begin{array}{l}0.5(2556)+(-1)(2742)+0.5(3250)\\ -9.5(2556)+18(2742)+(-8.5)(3250)\\ 45(2556)+(-80)(2742)+36(3250)\end{array}\right]\\ X=\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\left[\begin{array}{l}161\\ -2551\\ 12660\end{array}\right]\end{array}$

The solutions are $a=161,b=-2551,c=12660$ .

Thus, the system of matrix equations are $y=161t^2-2551t+12660$ .

c.

To determine

The graph of the parabola by using graphing utility.

Answer to Problem 72E

  

Explanation of Solution

Given information:

The table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

  $(c)$ Use the graphing utility to graph the parabola with the data.

Calculation:

From the given information,the table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

The graph of the parabola are as follows:

  

d.

To determine

The graph of the parabola by using graphing utility.

Answer to Problem 72E

No, because the model will increase without bound for higher values of $t$ .

Explanation of Solution

Given information:

The table shows the numbers of international travelers $y$ (in thousands) to the United States from South America from $2008$ through $2010$ .

  

  $(d)$ Do you believe the model is reasonable predictor of future numbers of travelers? Explain?

Calculation:

No, because the model will increase without bound for higher values of $t$ .