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erma4kov [3.2K]
2 years ago
11

Someone please help me with number 5 plz and thank you :)

Mathematics
2 answers:
Marysya12 [62]2 years ago
6 0

Answer:

75 Boys

HOPE THIS HELPS

-Todo <3

Step-by-step explanation:

60/4=15

15*5=75

kow [346]2 years ago
5 0

Answer:

75

Step-by-step explanation:

60/4=15

15*5=75

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Clara has 1,896 milliliters of juice to make punch. She needs to pour an equal amount of juice into 8 containers. How much juice
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Answer:

en cada vaso debe verter 237 mililitros

Step-by-step explanation:

1896 Divídelo en 8 es igual que 237 por 8

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2 years ago
Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob
Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that \mu = 36, \sigma = 5.5

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that n = 9. So

s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

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

By the Central Limit Theorem

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

Z = \frac{38 - 36}{1.8333}

Z = 1.09

Z = 1.09 has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

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

Z = \frac{34.5 - 36}{1.8333}

Z = -0.82

Z = -0.82 has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

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at the bank, Brent exchanges $50 in bills for 50 one-dollar coins. the total mass of the coins is 405 grams. Estimate the mass o
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Answer:

8.1 grams

Step-by-step explanation:


4 0
3 years ago
.....help me please
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Your answer would be A
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