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Stella [2.4K]
3 years ago
9

Find the minimum possible perimeter of a rectangular pentagon whose side is 15.0cm​

Mathematics
1 answer:
vekshin13 years ago
3 0

Answer:

75cm

Step-by-step explanation:

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Find the future value of an annuity due of $7,000 paid at the beginning of each 6-month period for 8 years if the interest rate
Oliga [24]

Answer: Im not sure but i think its 120,960

Step-by-step explanation: i know that there is 2 of the 6-month periods in a year and multiplied that by 8 and got 16. Then i did 7000x.08=560 and then i added that to 7000 and got 7560. FInally i multiplied that by 16 and got 120,960.

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3 years ago
The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye
trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

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}}.

In this problem, we have that:

\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

So

X = 500

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

By the Central Limit Theorem

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

Z = \frac{500 - 432}{114.16}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257.

X = 250

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

By the Central Limit Theorem

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

Z = \frac{250 - 432}{114.16}

Z = -1.59

Z = -1.59 has a pvalue of 0.0559.

So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.

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3 years ago
I really stuck on these questions can someone help
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