I need help on 6. Please show your work

Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters a

Agata [3.3K]

Answer:

The answers to the question is

(a) Jamie is gaining altitude at 1.676 m/s

(b) Jamie rising most rapidly at t = 15 s

At a rate of 2.094 m/s.

Step-by-step explanation:

(a) The time to make one complete revolution = period T = 15 seconds

Here will be required to develop the periodic motion equation thus

One complete revolution = 2π,

therefore the we have T = 2π/k = 15

Therefore k = 2π/15

The diameter = radius of the wheel = (diameter of wheel)/2 = 5

also we note that the center of the wheel is 6 m above ground

We write our equation in the form

y = $5*sin(\frac{2*\pi*t}{15} )+6$

When Jamie is 9 meters above the ground and rising we have

9 = $5*sin(\frac{2*\pi*t}{15} )+6$ or 3/5 = $sin(\frac{2*\pi*t}{15} )$ = 0.6

which gives sin⁻¹(0.6) = 0.643 = $\frac{2*\pi*t}{15}$

from where t = 1.536 s

Therefore Jamie is gaining altitude at

$\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =$ 1.676 m/s.

(b) Jamie is rising most rapidly when the velocity curve is at the highest point, that is where the slope is zero

Therefore we differentiate the equation for the velocity again to get

$\frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15})$ =0, π, 2π

Therefore $-sin(\frac{2\pi t}{15} )$ = 0 whereby t = 0 or

$\frac{2\pi t}{15}$ = π and t = 7.5 s, at 2·π t = 15 s

Plugging the value of t into the velocity equation we have

$\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =$ - 2/3π m/s which is decreasing

so we try at t = 15 s and we have $\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pi$ m/s

Hence Jamie is rising most rapidly at t = 15 s

The maximum rate of Jamie's rise is 2/3π m/s or 2.094 m/s.

7 0

3 years ago

Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to

attashe74 [19]

Answer:

a) Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

b) $X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c) $t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

$p_v =P(t_{(49)}>2.81)=0.0035$

d) If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X=32.12$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =30.2$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

$X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c. Perform the test and calculate P-value

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(49)}>2.81)=0.0035$

d. State your conclusion.

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

6 0

4 years ago

A music store buys instruments and then sells them for 30 % more than they paid

astra-53 [7]

Daddy remember when you made my butt sore

6 0

3 years ago

a sixth grade class has 12 boys and 24 girls consider this statement for every 2 boys there are 4 girls do you agree with this s

tigry1 [53]

This statement is correct as 12 is half of 24 and 2 is half of 4. We can also put this as the fraction 1/2 or 0.5 or 50% or 1:2. Therefore for every one bot there are two girls in the class.

5 0

3 years ago

Read 2 more answers

What is the parallel slope of m= -2

Anna35 [415]

You mean what is the slope of a line parallel to the line whose slope is -2?

The answer is -2... the slopes of parallel lines are equal

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3 years ago