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prisoha [69]
3 years ago
6

Bobs family of three was driving to Washington D.C.. They were going to stay overnight, sightsee during the next day, and return

home in the evening. They had to pay for dinner,breakfast,and lunch.they were to sleep at grandmas house. Breakfast at McDonalds was $4.89 each. Lunch at Kentucky Fried Chicken was $5 each. Dinner at Wendy's was $8.49 each. Was $60 enough money to pay for their food? Explain
Mathematics
1 answer:
ss7ja [257]3 years ago
5 0
Hello!

To find out if this food is enough we will add together all of the prices and multiply it by three to see if it is enough.

3(4.89+5+8.49)=55.14

Yes! They will have enough money with $4.86 to spare!

I hope this helps!
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Answer: You spent 5 hours babysitting.

Step-by-step explanation:

This situation can be represented by the  equation  y = 8x+ 10  where y is the money you earn and x is the  number of hours.

So if  y is equation to fifty dollars then what is x ?

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According to 2013 report from Population Reference Bureau, the mean travel time to work of workers ages 16 and older who did not
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Answer:

a) 48.80% probability that his travel time to work is less than 30 minutes

b) The mean is 30.7 minutes and the standard deviation is of 3.83 minutes.

c) 13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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:

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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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 30.7, \sigma = 23

a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?

This is the pvlaue of Z when X = 30. So

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

Z = \frac{30 - 30.7}{23}

Z = -0.03

Z = -0.03 has a pvalue of 0.4880.

48.80% probability that his travel time to work is less than 30 minutes

b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.

n = 36

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c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?

This is 1 subtracted by the pvalue of Z when X = 35. So

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

By the Central Limit Theorem

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

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Z = 1.12

Z = 1.12 has a pvalue of 0.8687

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