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Finger [1]
3 years ago
13

Marisa's car gets an average of 28 miles per gallon of gas. She plans to drive 200 miles today and 220 miles tomorrow. How many

gallons of gas should she expect to use in all?
A) 15 gallons
B) 28 gallons
C) 56 gallons
D) 67 gallons
Mathematics
1 answer:
nadezda [96]3 years ago
5 0

Answer:

15

Step-by-step explanation:

Add the total of miles Marisa is going to drive; 200 + 220 = 420

Next, divide the total miles by the average miles Marisa can travel on one gallon of gas; 420 / 28 = 15

(Hope this helps!)

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Assume that the population of human body temperatures has a mean of 98.6 degrees F and a standard deviation of 0.62 degrees F. I
dimulka [17.4K]

Answer:

0% probability of getting a mean temperature of 98.2 degrees F or lower.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

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:

\mu = 98.6, \sigma = 0.62, n = 106, s = \frac{0.62}{\sqrt{106}} = 0.06

Find the probability of getting a mean temperature of 98.2 degrees F or lower.

This is the pvalue of Z when X = 98.2. So

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

Z = \frac{98.2 - 98.6}{0.06}

Z = -6.67

Z = -6.67 has a pvalue of 0.

So there is a 0% probability of getting a mean temperature of 98.2 degrees F or lower.

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