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Answer :
We begin by organizing the data into groups that might represent one cycle. The table gives the number of cars recorded each hour for 12 hours. Notice that a repeating pattern is visible if we group the data in blocks of 3 hours. Thus, we try grouping hours as follows:
[tex]$$
\begin{array}{c|ccc}
\text{Cycle} & \text{Hour Values} & \text{Car Counts} \\
\hline
1 & 1-3 & 52,\;76,\;90 \\
2 & 4-6 & 75,\;91,\;104 \\
3 & 7-9 & 89,\;105,\;119 \\
4 & 10-12 & 103,\;121,\;135 \\
\end{array}
$$[/tex]
For each cycle, we determine the minimum and maximum number of cars, then compute the amplitude of that cycle as half the difference between the maximum and minimum values.
1. Cycle 1 (Hours 1--3):
- Minimum value: [tex]$52$[/tex]
- Maximum value: [tex]$90$[/tex]
The amplitude is:
[tex]$$
\frac{90 - 52}{2} = \frac{38}{2} = 19.
$$[/tex]
2. Cycle 2 (Hours 4--6):
- Minimum value: [tex]$75$[/tex]
- Maximum value: [tex]$104$[/tex]
The amplitude is:
[tex]$$
\frac{104 - 75}{2} = \frac{29}{2} = 14.5.
$$[/tex]
3. Cycle 3 (Hours 7--9):
- Minimum value: [tex]$89$[/tex]
- Maximum value: [tex]$119$[/tex]
The amplitude is:
[tex]$$
\frac{119 - 89}{2} = \frac{30}{2} = 15.
$$[/tex]
4. Cycle 4 (Hours 10--12):
- Minimum value: [tex]$103$[/tex]
- Maximum value: [tex]$135$[/tex]
The amplitude is:
[tex]$$
\frac{135 - 103}{2} = \frac{32}{2} = 16.
$$[/tex]
Although the amplitudes vary slightly from cycle to cycle due to an overall rising trend, in all cases the period is clearly [tex]$3$[/tex] hours and the amplitudes are around [tex]$15$[/tex].
Since each cycle covers 3 hours and the overall amplitude is approximately [tex]$15$[/tex], we conclude that the data set is approximately periodic with a period of [tex]$3$[/tex] hours and an amplitude of about [tex]$15$[/tex].
[tex]$$
\begin{array}{c|ccc}
\text{Cycle} & \text{Hour Values} & \text{Car Counts} \\
\hline
1 & 1-3 & 52,\;76,\;90 \\
2 & 4-6 & 75,\;91,\;104 \\
3 & 7-9 & 89,\;105,\;119 \\
4 & 10-12 & 103,\;121,\;135 \\
\end{array}
$$[/tex]
For each cycle, we determine the minimum and maximum number of cars, then compute the amplitude of that cycle as half the difference between the maximum and minimum values.
1. Cycle 1 (Hours 1--3):
- Minimum value: [tex]$52$[/tex]
- Maximum value: [tex]$90$[/tex]
The amplitude is:
[tex]$$
\frac{90 - 52}{2} = \frac{38}{2} = 19.
$$[/tex]
2. Cycle 2 (Hours 4--6):
- Minimum value: [tex]$75$[/tex]
- Maximum value: [tex]$104$[/tex]
The amplitude is:
[tex]$$
\frac{104 - 75}{2} = \frac{29}{2} = 14.5.
$$[/tex]
3. Cycle 3 (Hours 7--9):
- Minimum value: [tex]$89$[/tex]
- Maximum value: [tex]$119$[/tex]
The amplitude is:
[tex]$$
\frac{119 - 89}{2} = \frac{30}{2} = 15.
$$[/tex]
4. Cycle 4 (Hours 10--12):
- Minimum value: [tex]$103$[/tex]
- Maximum value: [tex]$135$[/tex]
The amplitude is:
[tex]$$
\frac{135 - 103}{2} = \frac{32}{2} = 16.
$$[/tex]
Although the amplitudes vary slightly from cycle to cycle due to an overall rising trend, in all cases the period is clearly [tex]$3$[/tex] hours and the amplitudes are around [tex]$15$[/tex].
Since each cycle covers 3 hours and the overall amplitude is approximately [tex]$15$[/tex], we conclude that the data set is approximately periodic with a period of [tex]$3$[/tex] hours and an amplitude of about [tex]$15$[/tex].
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