Answer :

The correct option is C. periodic with period of [tex]4[/tex] and amplitude of about [tex]15[/tex].

A periodic function can be defined as a type of function that has its values being repeated at regular interval of time (period).

From the image, we can deduce that the values are being repeated at regular interval of four ([tex]4[/tex]). Thus, its period is equal to four ([tex]4[/tex]).

For the amplitude:

[tex]Amplitude =\frac{38-9}{2}[/tex]

[tex]Amplitude = 14.5[/tex]

[tex]Amplitude = 15[/tex]

The complete question is

Is the data set approximately periodic?

If so, what are its period and amplitude?

A. not periodic

B. periodic with period of 4 and amplitude of about 7.5

C. periodic with period of 4 and amplitude of about 15

D. periodic with period of 5 and amplitude of about 15

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Rewritten by : Barada

Answer:

Option C. Periodic

Step-by-step explanation:

This problem is set with the number of days passing by versus the book borrowed, possibly by a library.

In order to see this, let's recall the concept of period and amplitude. The period in this case would be how much a number or a pattern is repeated over time. The amplitude would be the result of the difference between the maximum value and the minimum value of the table, in this case, the number of books borrowed.

Now, if you watch closely the table, we can watch a pattern. Which pattern? well, let's divide the table in 3 parts.

The first 4 days, we have: 102. 76, 91, 140

The second 4 days: 98, 74, 88, 137

The last 4 days: 100, 75, 92, 139

As you can see, every 4 days we get a pattern of books borrowed and almost the same number. [In day 1, 102 books were borrowed, and in day 5, 98 books were borrowed. The difference between that and the first day is only 4 books.

Day 2, was 76 books and day 6 was 74, again a very small difference. If we do the same for the remaining day we can conclude that every 4 days, the number of books borrowed follows a pattern, so we can conclude that the period would be 4.

Now to get the amplitude, let's see the average of difference between the maximun and minimum value:

A = 140 - 74 / 2 = 33.

As we can see, the value it's not exactly 25, but it's close to 33, and the problem states if the data set is approximately periodic, so, we can go with this option.