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2 years ago

2,652 divided by 48.

1 answer:

7 0

48 goes in to 2,652 how many times...that your answer...48 cannot go into 2 6 or 5 but it can into 265 5 times which equal240 subtract that from265 =25 ,48 cannot go in to 25 so bring down the 2 to make it 252 then divide 48 by 252 which goes 5 times equaling12 48 cannot go in to 12 so add a 0 to 2652 to make it 26520 then bring the 0 down to make 12 120 then divide 48 into that which it goes 2 times equaling 96 120subtract 96=24 48 cannot go into 24 so add another 0 to 26520 to make it 265200 bring down the zero to make 24 into 240 divide 48 into that which gives you240 subtract 240 from 240 = 0 so your answer is 48 divide into 2652=525

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The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the

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Using the normal distribution, we have that:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, the parameters are given as follows:

$\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358$

Hence:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{58 - 57}{22}$

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 57}{22}$

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

$Z = \frac{X - \mu}{s}$

$Z = \frac{58 - 57}{5.3358}$

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

$Z = \frac{X - \mu}{s}$

$Z = \frac{55 - 57}{5.3358}$

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

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mina [271]

Subtract b and then a = (zma) - b

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3 years ago