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Mila [183]
1 year ago
14

Just this please helppp

Mathematics
1 answer:
Murrr4er [49]1 year ago
5 0

The height of the <em>water</em> depth is h = 14 + 6 · sin (π · t/6 + π/2), where t is in hours, and the height of the Ferris wheel is h = 21 + 19 · sin (π · t/20 - π/2), where t is in seconds. Please see the image to see the figures.

<h3>How to derive equations for periodical changes in time</h3>

According to the two cases described in the statement, we have clear example of <em>sinusoidal</em> model for the height as a function of time. In this case, we can make use of the following equation:

h = a + A · sin (2π · t/T + B)     (1)

Where:

  • a - Initial position, in meters.
  • A - Amplitude, in meters.
  • t - Time, in hours or seconds.
  • T - Period, in hours or seconds.
  • B - Phase, in radians.

Now we proceed to derive the equations for each case:

Water depth (u = 20 m, l = 8 m, a = 14 m, T = 12 h):

A = (20 m - 8 m)/2

A = 6 m

a = 14 m

Phase

20 = 14 + 6 · sin B

6 = 6 · sin B

sin B = 1

B = π/2

h = 14 + 6 · sin (π · t/6 + π/2), where t is in hours.

Ferris wheel (u = 40 m, l = 2 m, a = 21 m, T = 40 s):

A = (40 m - 2 m)/2

A = 19 m

a = 21 m

Phase

2 = 21 + 19 · sin B

- 19 = 19 · sin B

sin B = - 1

B = - π/2

h = 21 + 19 · sin (π · t/20 - π/2), where t is in seconds.

Lastly, we proceed to graph each case in the figures attached below.

To learn more on sinusoidal models: brainly.com/question/12060967

#SPJ1

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Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

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Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

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H₁: μ₁ > μ₂

α: 0.05

t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66

t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

I hope this helps!

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