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julsineya [31]
4 years ago
5

Alyssa received $30,000 from an inheritance and invests part in a certificate of deposit (CD) paying 5% annual interest and the

rest in Apple Inc. offering an annual return of 11%. If the annual income from her investments is $2,220, how much did Alyssa invest in each?
a) How much money is Alyssa investing?

b) How long is Alyssa investing the money?

c) What are the two investments and corresponding interest rates?
_______and _______

d) What are you asked to determine
Mathematics
1 answer:
sammy [17]4 years ago
5 0
B would be the mostikeley answer
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(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To m
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Answer:

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

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 manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that \mu = 50.

We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?

We have to find the pvalue of Z when X = 47.3.

We are working with a sample mean, so we use the standard deviation of the sample in the place of \sigma. That is s = 9

So

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

Z = \frac{47.3-50}{9}

Z = -0.3

Z = -0.3 has a pvalue of 0.3821.

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

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