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Answer :
We start with the data for cups of coffee over 12 days:
[tex]$$
\begin{array}{|c|cccccccccccc|}
\hline
\text{Day} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\text{Cups} & 16 & 30 & 8 & 14 & 28 & 10 & 15 & 31 & 11 & 14 & 29 & 9 \\
\hline
\end{array}
$$[/tex]
A periodic pattern is suspected, and the data seems to repeat every 4 days. To verify this, we group the data based on a period of 4:
- Group 1 (Days 1, 5, 9):
Values: [tex]$16$[/tex], [tex]$28$[/tex], and [tex]$11$[/tex].
- Group 2 (Days 2, 6, 10):
Values: [tex]$30$[/tex], [tex]$10$[/tex], and [tex]$14$[/tex].
- Group 3 (Days 3, 7, 11):
Values: [tex]$8$[/tex], [tex]$15$[/tex], and [tex]$29$[/tex].
- Group 4 (Days 4, 8, 12):
Values: [tex]$14$[/tex], [tex]$31$[/tex], and [tex]$9$[/tex].
For each group, the amplitude is approximated as half the difference between the maximum and minimum values.
1. Group 1:
Maximum = [tex]$28$[/tex], Minimum = [tex]$11$[/tex]
Amplitude:
[tex]$$
\frac{28 - 11}{2} = \frac{17}{2} = 8.5
$$[/tex]
2. Group 2:
Maximum = [tex]$30$[/tex], Minimum = [tex]$10$[/tex]
Amplitude:
[tex]$$
\frac{30 - 10}{2} = \frac{20}{2} = 10.0
$$[/tex]
3. Group 3:
Maximum = [tex]$29$[/tex], Minimum = [tex]$8$[/tex]
Amplitude:
[tex]$$
\frac{29 - 8}{2} = \frac{21}{2} = 10.5
$$[/tex]
4. Group 4:
Maximum = [tex]$31$[/tex], Minimum = [tex]$9$[/tex]
Amplitude:
[tex]$$
\frac{31 - 9}{2} = \frac{22}{2} = 11.0
$$[/tex]
Next, we compute the average amplitude from these four groups:
[tex]$$
\text{Average Amplitude} = \frac{8.5 + 10.0 + 10.5 + 11.0}{4} = \frac{40}{4} = 10.0
$$[/tex]
The data is approximately periodic with a period of [tex]$4$[/tex] days and an average amplitude of about [tex]$10$[/tex] cups.
Thus, the correct answer is:
[tex]$$\textbf{Periodic with a period of 4 and an amplitude of about 10.}$$[/tex]
[tex]$$
\begin{array}{|c|cccccccccccc|}
\hline
\text{Day} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\text{Cups} & 16 & 30 & 8 & 14 & 28 & 10 & 15 & 31 & 11 & 14 & 29 & 9 \\
\hline
\end{array}
$$[/tex]
A periodic pattern is suspected, and the data seems to repeat every 4 days. To verify this, we group the data based on a period of 4:
- Group 1 (Days 1, 5, 9):
Values: [tex]$16$[/tex], [tex]$28$[/tex], and [tex]$11$[/tex].
- Group 2 (Days 2, 6, 10):
Values: [tex]$30$[/tex], [tex]$10$[/tex], and [tex]$14$[/tex].
- Group 3 (Days 3, 7, 11):
Values: [tex]$8$[/tex], [tex]$15$[/tex], and [tex]$29$[/tex].
- Group 4 (Days 4, 8, 12):
Values: [tex]$14$[/tex], [tex]$31$[/tex], and [tex]$9$[/tex].
For each group, the amplitude is approximated as half the difference between the maximum and minimum values.
1. Group 1:
Maximum = [tex]$28$[/tex], Minimum = [tex]$11$[/tex]
Amplitude:
[tex]$$
\frac{28 - 11}{2} = \frac{17}{2} = 8.5
$$[/tex]
2. Group 2:
Maximum = [tex]$30$[/tex], Minimum = [tex]$10$[/tex]
Amplitude:
[tex]$$
\frac{30 - 10}{2} = \frac{20}{2} = 10.0
$$[/tex]
3. Group 3:
Maximum = [tex]$29$[/tex], Minimum = [tex]$8$[/tex]
Amplitude:
[tex]$$
\frac{29 - 8}{2} = \frac{21}{2} = 10.5
$$[/tex]
4. Group 4:
Maximum = [tex]$31$[/tex], Minimum = [tex]$9$[/tex]
Amplitude:
[tex]$$
\frac{31 - 9}{2} = \frac{22}{2} = 11.0
$$[/tex]
Next, we compute the average amplitude from these four groups:
[tex]$$
\text{Average Amplitude} = \frac{8.5 + 10.0 + 10.5 + 11.0}{4} = \frac{40}{4} = 10.0
$$[/tex]
The data is approximately periodic with a period of [tex]$4$[/tex] days and an average amplitude of about [tex]$10$[/tex] cups.
Thus, the correct answer is:
[tex]$$\textbf{Periodic with a period of 4 and an amplitude of about 10.}$$[/tex]
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