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vlada-n [284]
3 years ago
10

Throughout the day, the water level, h(t), at Sunny Beach varies with the tides. Joe, a frequent visitor of Sunny Beach, arrived

at the beach at 12:00 p.m., and collected the following data, where t represents the number of hours since Joe arrived at the beach. t 0 1 2 3 4 5 h(t) 5 4.41 3 1.59 1 1.59 Joe lost some of his data, but he knows that the highest tide occurs at 12:00 p.m., and he was able to use a trigonometric function to model the height of the tide at any time. What is the period of the function that can be used to model the height of the tide

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
5 0

Answer:

8 hours

Step-by-step explanation:

Joe's data shows the minimum to be 4 hours after his arrival at the beach. The time between maximum and minimum is half a period. (The other half brings the function back to its maximum.) If half a period is 4 hours, the whole period is 8 hours.

Svetlanka [38]3 years ago
4 0

Answer:

Around 8 hours so D

Step-by-step explanation:

Just a normal graph checked on 3 websites all say the same

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It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
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Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

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The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

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