Chapter 3, Problem 47CLT

a.

To determine

To find: A quadratic model, an exponential model and a power model for given data and then to identify the coefficient of determination for each model.

Answer to Problem 47CLT

Quadratic model is $S=-1.17t^2+40.23t+147.85$ , its coefficient of determination is about 1y1y3 $0.9905$ .

Exponential model is $S=229.068{(1.06523)}^t$ , its coefficient of determination is about $0.9486$ .

Power model is

  $S=160.68{(t)}^{0.42942}$ , its coefficient of determination is about $0.9921$ .

Explanation of Solution

Given Information: A relevant table and $t=3$ corresponds to $2003$ .

Calculation:

Using regression feature of graphing utility to calculate a quadratic model,

  

  $\Rightarrow$ Quadratic model is $S=-1.17t^2+40.23t+147.85$ , its coefficient of determination is about $0.9905$ .

Using regression feature of graphing utility to calculate an exponential model,

  

  $\Rightarrow$ Exponential model is $S=229.068{(1.06523)}^t$ , its coefficient of determination is about $0.9486$ .

Using regression feature of graphing utility to calculate a power model,

  

  $\Rightarrow$ Power model is $S=160.68{(t)}^{0.42942}$ , its coefficient of determination is about $0.9921$ .

b.

To determine

To graph: Each model with the original data using a graphing utility.

Explanation of Solution

Given Information: A relevant table and models from previous part.

Graphs:

Quadratic model:

  

Exponential model:

  

Power model:

  

c.

To determine

To find: The model that best fits the data.

Answer to Problem 47CLT

Power model best fits the data because of its favorable coefficient of determination.

Explanation of Solution

Given Information: Models and their coefficient of determination (from previous parts).

Calculation:

Out of three values, coefficient of determination $(r^2)$ that is most closest to $1$ is $0.9921$ which corresponds to the power model, therefore, power model best fits the data.

Thus, power model best fits the data because of its favorable coefficient of determination.

d.

To determine

To predict: The annual sales of Wal-Mart in $2020$ and then to write if the answer is reasonable.

Answer to Problem 47CLT

In $2020$ , annual sales would be about $\$581.63billion$ , it is reasonable as with time, demand increases and sales also increase.

Explanation of Solution

Given Information: Power model calculated previously.

Calculation:

In previous part, power model was chosen because it best fits the data, using power model to predict for $t=20$ ,

  $\begin{array}{l}$ S=160.68{(t)}^{0.42942}$$ =160.68{(20)}^{0.42942}$=581.63\end{array}$

  $\Rightarrow$ In $2020$ , annual sales would be near about $\$581.63billion$ .

This is reasonable because as population and awareness increase, demand also increases and as a result sales soar.

Thus, in $2020$ , annual sales would be about $\$581.63billion$ , it is reasonable as with time customers increase and sales also increase.