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Alchen [17]
3 years ago
7

Ive been at this for abt 30 min. help?

Mathematics
1 answer:
gregori [183]3 years ago
6 0

Answer:

1+\frac{-63}{x^2-1}

Step-by-step explanation:

If you foil this equation you will get:

\frac{(x+8)(x-8)}{(x+1)(x-1)}

And by dividing it, you will get;

1+\frac{-63}{x^2-1}

That is your answer!

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The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 104 inches, and a standard
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Answer:

91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

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 normally distributed samples are 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:

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The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 104, \sigma = 14, n = 49, s = \frac{14}{\sqrt{49}} = 2

What is the probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

This is the pvalue of Z when X = 106.8. So

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

By the Central Limit Theorem

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

Z = \frac{106.8 - 104}{2}

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0.9192 = 91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

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