Chapter 11.4, Problem 60E

a.

To determine

To create: a scatter plot of data and graph the model in the same viewing window and how do they compare.

Answer to Problem 60E

Explanation of Solution

Given information: A model for the data is given by:

  $\mathrm{S}(\mathrm{t})=\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2},6\le \mathrm{t}\le 17$.

Where t represents the year, with t = 6 corresponding to 2006.

Calculation:

The graph of the model is shown below.

  

b.

To determine

To predict: the mean salaries in 2018 and 2019.

Answer to Problem 60E

  $\mathrm{S}(18)=149812\text{dollars}$ .

  $\mathrm{S}(19)=157212\text{dollars}$ .

Explanation of Solution

Given information: $\mathrm{S}(\mathrm{t})=\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2},6\le \mathrm{t}\le 17$

Calculation:

  $\begin{array}{l}\mathrm{S}(\mathrm{t})=\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2}\\ 2018\Rightarrow \mathrm{t}=18\Rightarrow \mathrm{S}(18)=\frac{-201.37+240.816(18)-7.3976{(18)}^2}{1+1.861(18)-0.0707{(18)}^2}=148.812\text{thousand}\text{dollar}\\ \mathrm{S}(18)=149812\text{dollars}\text{.}\\ 2019\Rightarrow \mathrm{t}=19\Rightarrow \mathrm{S}(19)=\frac{-201.37+240.816(19)-7.3976{(19)}^2}{1+1.861(19)-0.0707{(19)}^2}=157.212\text{thousand}\text{dollar}\\ \mathrm{S}(19)=157212\text{dollars}\end{array}$

c.

To determine

To find: the limit of the model as $\mathrm{t}\to \mathrm{\infty }$ and interpret its meaning in the context of the situation.

Answer to Problem 60E

As time passes, the mean annual salaries of a financial manager in United States approach 0.

Explanation of Solution

Given information: $\mathrm{S}(\mathrm{t})=\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2},6\le \mathrm{t}\le 17$

Calculation:

  $\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\mathrm{S}(\mathrm{t})=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2}=0$

As time passes, the mean annual salaries of a financial manager in United States approach 0.

d.

To determine

To find: is this a good model for predicting the mean annual salaries in future years.

Answer to Problem 60E

No, this model is not a good model for predicting the mean annual salaries in future years because $\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\mathrm{S}(\mathrm{t})=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2}=0$.

Explanation of Solution

Given information:

Calculation:

No, this model is not a good model for predicting the mean annual salaries in future years because $\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\mathrm{S}(\mathrm{t})=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{-201.37+240.816\mathrm{t}-7.3976{\mathrm{t}}^2}{1+1.861\mathrm{t}-0.0707{\mathrm{t}}^2}=0$.