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exis [7]
3 years ago
8

Mohammed buys 31 cups of coffee each week. If a cup of coffee costs d dollars and Mohammed buys approximately the same number of

coffees every day, which would be a mathematical way to express the amount of money that Mohammed spends on coffee each day?
Mathematics
2 answers:
choli [55]3 years ago
5 0
Mohammed spends 31(d) on coffee each day
Advocard [28]3 years ago
5 0
31/7=4 4*d=amount spent a day on coffee
(7 for the days of the week)
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Solids of rotation will always have a curved face, so they’re always non-polyhedral.

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The Office of Student Services at a large western state university maintains information on the study habits of its full-time st
Vera_Pavlovna [14]

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0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

Step-by-step explanation:

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

Normal Probability Distribution:

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Central Limit Theorem

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

Mean of 20 hours, standard deviation of 6:

This means that \mu = 20, \sigma = 6

Sample of 150:

This means that n = 150, s = \frac{6}{\sqrt{150}}

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}

Z = 2.04

Z = 2.04 has a p-value of 0.9793

X = 19.5

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

Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}

Z = -1.02

Z = -1.02 has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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