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babymother [125]
3 years ago
10

Help simply. this math problem

Mathematics
2 answers:
saveliy_v [14]3 years ago
4 0

- 4 1/5 = -21/5 = -42/10

-13 1/10 = -131/10

-42/10 -  -131/10=

-42/10 + 131/10 = 89/10 = 8 9/10

 answer is A

tatuchka [14]3 years ago
4 0
<span><span><span>−42</span>/10</span>+<span>131/10</span></span><span>=<span><span><span>−42</span>/10</span>+<span>131/10</span></span></span><span>=<span><span><span>−42</span>+131</span>/10</span></span><span>=<span>8910</span></span><span>(Decimal: 8.9)</span>Answer: A
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Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

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At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

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The test statistic is:

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28.76 is tested at the null hypothesis:

This means that \mu = 28.76

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

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z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level \alpha = 0.05.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

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The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

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