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ziro4ka [17]
3 years ago
11

Which equation determines the proportion of ground squirrels alive at the start of year 1-2?

Mathematics
1 answer:
Mrrafil [7]3 years ago
7 0

Hi! If you took the same test as me, the answer should be "64/80 = 0.800"

But, you might have a different graph.

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Step-by-step explanation:

180 - 100 = 80

80 = 30 = 110

180 - 110 = 70

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The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontradit
shutvik [7]

Answer:

There is a 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.

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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

For this problem, we have that:

Based on the study results, we can assume the population mean and standard deviation for the GPA of nontraditional students is \mu = 3.5 and \sigma = 0.5.

We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of \sigma in the z score formula.

s = \frac{\sigma}{\sqrt{100}} = \frac{0.5}{10} = 0.05.

What is the probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65?

This is 1 subtracted by the pvalue of Z when X = 3.65. So

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

Z = \frac{3.65 - 3.50}{0.05}

Z = 3

A zscore of 3 has a pvalue of 0.9987.

So, there is a 1-0.9987 = 0.0013 = 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.

4 0
3 years ago
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