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Svetlanka [38]
3 years ago
15

PLEASE ANSWER THIS QUESTION RIGHT PLEASE Devante Has a lunch account in the school cafeteria. His starting balance at the beginn

ing of the month is $35.50 The first week Devante But three lunches and his account balance was$28. The second week Devante  Bought five lunches and his account balance was $15.50 in a model that relates the balance of his lunch account as a function of the number of lunches he buys what does the rate of change represent? THE ANSWER CHOICES ARE IN THE PICTURE ABOVE

Mathematics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

C) The amount of money in his lunch account

Step-by-step explanation:

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A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11. Use the TI-84 Plus calculator. (a) I
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Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

b) There is a 15.97% probability that x will be between 54 and 55.

c) The 47th percentile of x is X = 52.877.

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.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that \mu = 53.

We have to find the standard deviation of the sample, that is:

\sigma = \frac{11}{\sqrt{45}} = 1.64

(a) Is it appropriate to use the normal distribution to find probabilities for x?

We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

(b) Find the probability that x will be between 54 and 55.

This is the pvalue of the Z score when X = 55 subtracted by the pvalue of the Z score when X = 54.

X = 55

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

Z = \frac{55 - 53}{1.64}

Z = 1.22

Z = 1.22 has a pvalue of 0.88877.

X = 54

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

Z = \frac{54 - 53}{1.64}

Z = 0.61

Z = 0.61 has a pvalue of 0.72907.

So, there is a 0.88877 - 0.72907 = 0.1597 = 15.97% probability that x will be between 54 and 55.

(c) Find the 47th percentile of x. Round the answer to at least two decimal places.

This is the value of X when Z has a pvalue of 0.47;

This is between Z = -0.07 and Z = -0.08. So we use Z = -0.075.

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

-0.075 = \frac{X - 53}{1.64}

X = 52.877

The 47th percentile of x is X = 52.877.

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Step-by-step explanation:

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Step-by-step explanation:

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