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Lapatulllka [165]
4 years ago
13

SOME ONE PLEASE HELP Determine the sum: (9r3s2t − 4r2s2t2 − 6rs2t3) + (−4r3s2t + 8rs2t3).

Mathematics
1 answer:
mote1985 [20]4 years ago
4 0
The answer is
<span><span><span><span>5<span>r3</span></span><span>s2</span></span>t</span>−<span><span><span>4<span>r2</span></span><span>s2</span></span><span>t2</span></span></span>+<span><span><span>2r</span><span>s2</span></span><span>t<span>3</span></span></span>
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marshall27 [118]

Answer:

There is an 84% probability that your standard refrigerator will last between 12 and 20 years.

Step-by-step explanation:

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 standard refrigerator's mean life expectancy is 18 years with a standard deviation of 2 years. This means that \mu = 18, \sigma = 2.

What is the probability that your standard refrigerator will last between 12 and 20 years?

This probability is the pvalue of Z when X = 20 subtracted by the pvalue of Z when X = 12. So

X = 20

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

Z = \frac{20 - 18}{2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 12

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

Z = \frac{12- 18}{2}

Z = -3

Z = -3 has a pvalue of 0.0014

This means that there is a 0.8413-0.0014 = 0.8399 = 0.84 = 84% probability that your standard refrigerator will last between 12 and 20 years.

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Step-by-step explanation:

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Step-by-step explanation:

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