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kramer
3 years ago
15

Which inequality is true for 50pts​

Mathematics
2 answers:
shepuryov [24]3 years ago
8 0

Answer:

B. (5+2)^{2} > 5^2 + 2^2

Step-by-step explanation:

The inequality "B" is correct because when you simplify it, you can see that it holds true:

7^{2} > 25 + 4

49 > 29

49 is indeed greater than 29, making this inequality true.

enot [183]3 years ago
8 0

Answer:

B. (5+2)^{2} > 5^2 + 2^2

The inequality "B" is correct because when you simplify it, you can see that it holds true:

Step-by-step explanation:

7^{2} > 25 + 4

49 > 29

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Solve:<br>3 = -2v - V<br>V=​
Y_Kistochka [10]

Answer:

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Step-by-step explanation:

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Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa
Debora [2.8K]

Answer:

a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

b) 0.4129 = 41.29% probability that the mean return will be less than 8%

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 8.7% and standard deviation 20.2%.

This means that \mu = 8.7, \sigma = 20.2

40 years:

This means that n = 40, s = \frac{20.2}{\sqrt{40}}

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?

This is 1 subtracted by the pvalue of Z when X = 13. So

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

By the Central Limit Theorem

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

Z = \frac{13 - 8.7}{\frac{20.2}{\sqrt{40}}}

Z = 1.35

Z = 1.35 has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

(b) What is the probability that the mean return will be less than 8%?

This is the pvalue of Z when X = 8. So

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

Z = \frac{8 - 8.7}{\frac{20.2}{\sqrt{40}}}

Z = -0.22

Z = -0.22 has a pvalue of 0.4129

0.4129 = 41.29% probability that the mean return will be less than 8%

8 0
2 years ago
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