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DerKrebs [107]
4 years ago
5

A light plane can fly at a rate of 100 mph in calm air. Traveling with the plane flew 360 mi in the same amount of time as it fl

ew 240 mi against the wind. Find the rate of the wind.
Mathematics
1 answer:
Lynna [10]4 years ago
7 0

Answer:

The rate of wind is 20 mph.

Step-by-step explanation:

The actual speed of the plane is 100 mph.

Now, it is given that, the plane takes the same time to travel 360 miles with the flow of air and to travel 240 miles against the flow of air.

Let the equal time is t hrs.

If, the speed of the plane is u and that of wind is v then we can write,

360/t =u+v ....(1) and  

240/t =u-v .....(2)

Now, adding (1) and (2), we get, 2u =600/t, ⇒ 2×100 =600/t, ⇒t = 3 hours.

Hence, from equation (2), we get v = u-240/t, ⇒ v =100-240/3 =20 mph

Therefore, the rate of wind is 20 mph. (Answer)

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Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation  

n=2341 represent the random sample taken    

X represent the people that they have watched digitally streamed TV programming on some type of device

\hat p=0.55 estimated proportion of people that they have watched digitally streamed TV programming on some type of device  

\alpha=0.01 represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval  

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

p \sim N (p, \sqrt{\frac{p(1-p)}{n}}

Part a) Confidence interval

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523

0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

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Answer:

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63 mph is 13 mph more than 50 mph, so the fine will be that for 13 mph over the speed limit. The fine corresponding to 13 mph over the limit is $200.

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