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3 years ago

13

You are going to rent a car to drive while you are on vacation. There are two car rental packages for you to choose from. The fi

st package offers a $120 per day flat rate. The second package offers a $100 flat rate plus $0.55 per mile. How many miles would you need to drive while on vacation in order to make the first package the better deal? A.24 miles B. 28 miles C . 32 miles D.37 miles E.45 miles

1 answer:

dangina [55]3 years ago

7 0

Answer:

E. 45 miles

Step-by-step explanation:

The difference between the packages is $20. You would need to drive at least 40 miles since 40(0.5) =$20. The best answer is E. 45 miles

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Answer:

a) 79.10% probability that a student had a score less than 320.

b) 46.63% probability that a student had a score between 250 and 300.

c) 99.25% of the students had a test score greater than 200

d) 350.865

e) 265.025

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 290, \sigma = 37$

a) Find the probability that a student had a score less than 320.

This is the pvalue of Z when X = 320. So

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

$Z = \frac{320 - 290}{37}$

$Z = 0.81$

$Z = 0.81$ has a pvalue of 0.7910

79.10% probability that a student had a score less than 320.

b) Find the probability that a student had a score between 250 and 300.

This is the pvalue of Z when X = 300 subtracted by the pvalue of Z when X = 250.

X = 300

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

$Z = \frac{300 - 290}{37}$

$Z = 0.27$

$Z = 0.27$ has a pvalue of 0.6064

X = 250

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

$Z = \frac{250 - 290}{37}$

$Z = -1.08$

$Z = -1.08$ has a pvalue of 0.1401

0.6064 - 0.1401 = 0.4663

46.63% probability that a student had a score between 250 and 300.

c) What percent of the students had a test score greater than 200?

This is 1 subtracted by the pvalue of Z when X = 200. So

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

$Z = \frac{200 - 290}{37}$

$Z = -2.43$

$Z = -2.43$ has a pvalue of 0.0075

1 - 0.0075 = 0.9925

99.25% of the students had a test score greater than 200

d) What is the lowest score that would still place a student in the top 5% of the scores?

X when Z has a pvalue of 1-0.05 = 0.95. So X when Z = 1.645.

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

$1.645 = \frac{X - 290}{37}$

$X - 290 = 37*1.645$

$X = 350.865$

e) What is the highest score that would still place a student in the bottom 25% of the scores

X when Z has a pvalue of 0.25. So X when Z = -0.675

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

$-0.675 = \frac{X - 290}{37}$

$X - 290 = 37*(-0.675)$

$X = 265.025$

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