Chapter 4.8, Problem 71E

a.

To determine

To create: a scatter plot of the data.

Explanation of Solution

Given information: The table shows the average sales S (in millions of dollars) of an

outerwear manufacture for each month t , where t = 1 represents January.

Month, t	Sales S
1	6.73
2	5.58
3	4.00
4	2.43
5	1.27
6	0.85
7	1.27
8	2.43
9	4.00
10	5.58
11	6.73
12	7.15

Calculation:

The scatter plot of the given table data is shown below.

  

b.

To determine

To find: a trigonometric model that fits the data and graph the model on your scatter plot and how well does the model fit the data.

Answer to Problem 71E

  $s=3.15\sin (0.524t+1.5(86))+4.$

The model fits the data well.

Explanation of Solution

Calculation:

Using the trigonometric model calculator the trigonometric model of the given data table is:

  $s=3.15\sin (0.524t+1.5(86))+4.$

The model fits the data well.

The graph of the above model is given below.

  

c.

To determine

To find: the period of the model and it is reasonable given the context, explain.

Answer to Problem 71E

  $\text{Period}=\frac{2\pi }{0.524}\approx 11.99\text{or}12\text{months}\text{.}$

The period of 12 months is measurable because outer wear sales would be seasonal.

Explanation of Solution

Given information: $s=3.15\sin (0.524t+1.5(86))+4.$

Calculation:

  $\text{Period}=\frac{2\pi }{0.524}\approx 11.99\text{or}12\text{months}\text{.}$

The period of 12 months is measurable because outer wear sales would be seasonal.

d.

To determine

To interpret: the meaning of the model’s amplitude in the context of the problem.

Answer to Problem 71E

The amplitude is 3.15 and gives the maximum displacement or greater and least average sales from 4 + 3.15 = $ 7.15 billion to 4 - 3.15= $ 0.85 billion.

Explanation of Solution

Given information: $s=3.15\sin (0.524t+1.5(86))+4.$

Calculation:

The amplitude is 3.15 and gives the maximum displacement or greater and least average sales from 4 + 3.15 = $ 7.15 billion to 4 - 3.15= $ 0.85 billion.