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drek231 [11]
2 years ago
10

An object that weighs 1 pound on earth weighs only 0.17 pounds on the moon. Zach's dog weighs 17 1/2 pounds on earth. How much m

ore does his dog weigh on earth than on the moon?
Mathematics
1 answer:
PtichkaEL [24]2 years ago
3 0
 its actually .16 not .17
 284.192
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The point (−3,15) undergoes a translation of 9 units right and 10
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6, 5

Step-by-step explanation:

-3 + 9 = 6

15-10 = 5

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Ashley Sigler gave 2/5 of her bonus check of $255 to a missionary family. How much
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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

#SPJ1

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