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umka2103 [35]
3 years ago
9

The volume of a right rectangular prism is 573.3 cm³. The length of the right rectangular prism is 7.35 cm and the height is 16.

25 cm. What is the width of the right rectangular prism? Enter your answer in the box.
Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0
The width of the right rectangular prism is 4.8. You multiply 7.35 by 16.25 and then you divide 573.3 by that answer. So, 573.3 ÷ 119.4375 = 4.8. Hope this helped
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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m
love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is c = 20.6153.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that \mu = 20.4, \sigma = 1.23

Consider the sample mean weight of 100 watermelons of this variety. This means that n = 100.

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when X = 20.27.

Since we are working with the sample mean, we use s instead of \sigma in the Z score formula

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

Z = \frac{20.27 - 20.4}{0.123}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of X = c that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use Z = 1.75

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

1.75 = \frac{c - 20.4}{0.123}

c - 20.4 = 0.123*1.75

c = 20.6153

This value is c = 20.6153.

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3 years ago
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I hope this helps you



x^2-7x-44

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Answer:

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Step-by-step explanation:

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8 is greater than 1 so you add one to 1 which equals 2, so your answer would be 0.22

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Answer:

Unfortunately it can't be solved.

Step-by-step explanation:

x is the all real number. which means it will have many solutions.

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