A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor? The mean, because the distribution of wind speeds is skewed. The median, because the distribution of wind speeds is symmetric. The mean, because the distribution of wind speeds is symmetric. The median, because the distribution of wind speeds is skewed. The proportion, because the distribution of wind speeds is skewed. The proportion, because the distribution of wind speeds is symmetric.

Answers

Answer 1

Answer:

9514 1404 393

Answer:

The median, because the distribution of wind speeds is skewed.

Step-by-step explanation:

A few higher speeds will raise the mean. For a skewed distribution, the median may be a more appropriate measure of center.

Additional comment

From an engineering point of view, one might have to consider speeds only within some range, as the wind turbines may be shut down for speeds too low or too high. Likely the power generated is not proportional to speed, so a weighted average (of the square of speed) would probably be more useful. Unfortunately, this question does not get into such subtleties.

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Answers

Answer:

x=W/10y

Step-by-step explanation:

If the first common multiple of two numbers is
6, find the fourth common multiple

Answers

24

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And to check it is safe to assume the two numbers are 2 and 3

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A brainliest would be appreciated!!

Hope this helped!

Answer:

The first common number are 2 and 3

their comnon multiple is 6

So, the fourth number will be 24 because

multiple means we should multiply 4*6 it will be 24.

Solve for the area of ΔABC to the nearest whole number. A) 13 cm2 B) 26 cm2 C) 53 cm2 D) 106 cm2

Answers

Answer:

The answer is 53 cm2 or c.

If G is midpoint of FH, find FG

Answers

The answer is 8 ihinghkki

What’s the correct answer for this question?

Answers

Answer:

Step-by-step explanation:

In the attached file

Answer:

Step-by-step explanation:

I put the answer in an atachement

In a survey of 1016 ?adults, a polling agency? asked, "When you? retire, do you think you will have enough money to live comfortably or not. Of the 1016 ?surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.A. There is a 99?% probability that the true proportion of worried adults is between ___ and ___.

B. 99?% of the population lies in the interval between ___ and ___.

C. There is 99?% confidence that the proportion of worried adults is between ___ and ___.

Answers

Answer:

C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Step-by-step explanation:

1) Data given and notation

n=1016 represent the random sample taken

X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement

$\hat p=(535)/(1016)=0.527$ estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic

p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement

2) Confidence interval

The confidence interval would be given by this formula

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

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:

$0.527 - 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.487$

$0.527 + 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.567$

And the 99% confidence interval would be given (0.487;0.567).

There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Final answer:

To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.

Explanation:

To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).

Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.

The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.

Learn more about Confidence Interval here:

brainly.com/question/34700241

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