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3 years ago

WILL MARK BRAINLEIST HURRY

Mathematics

2 answers:

Luden [163]3 years ago

6 0

Answer:

The IQR is the best measurement of spread for games and movies.

Step-by-step explanation:

I took the test

iren [92.7K]3 years ago

4 0

Answer:

Hello, The answer it's D.

Step-by-step explanation:

Assume that the data for both movies and basketball games are normally distributed.

Therefore, the median and the mean are assumed equal.

The standard deviation, σ, is related to the interquartile range by

IQR = 1.35

From the data, we can say the following:

Movies:

Range = 150 - 60 = 90 (approx)

Q1 = 62 (approx), first quartile

Q3 = 120 (approx), third quartlie

Q2 (median) = 90 (approx)

IQR = Q3 - Q1 = 58

σ = IQR/1.35 = 58/1.35 = 43

Basketball:

Range = 150 - 90 = 60 approx

Q1 = 95 (approx)

Q3 = 145 (approx)

Q2 = 125 (approx)

IQR = 145 - 95 = 50

σ = 50/1.35 = 37

Test the given answers.

A. The IQRs are approximately equal, so they are not good measures of spread. This is not a good answer.

B. The std. deviation is a better measure of spread for basketball. This is not a good answer.

C. IQR is not a better measure of spread for basketball games. This is not a good answer.

D. The standard deviation is a good measure of spread for both movies and basketball. This is the best answer.

Answer: D

Have a Great Day!

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liberstina [14]

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Perimeter=40ft

The perimeter of a rectangle is:

P=2l+2w

Since we know the length is 5 and the perimeter is 40, we can solve for the width.

P=2l+2w
40=2*(5)+2w
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subtract 10 from both sides
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divide both sides by 2
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The width of the fence is 15feet.

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3 years ago

What would the following equation look like in augmented matrix form? 2x+y=6 x-3y=-2

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3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

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What's the question?

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3 years ago

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solniwko [45]

Answer:

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Step-by-step explanation:

Since we see that sides AD and DC are equal, as well as sides BE and EC, we can use this information to conclude that the value of 2x-7 is equal to double the value of DE.

2x - 7 = 4(2)

2x - 7 = 8

2x - 7 + 7 = 8 + 7

2x = 15

x = 7.5

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