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leonid [27]
3 years ago
8

Pepe's pizza makes 52 pizza crusts the first week and 180 pizza crusts each subsequent week. Dianne's Delicatessen makes 26 pizz

a crusts the first week and 90 pizza crusts each subsequent week. In how many weeks will the total number of pizza crusts made by Pepe's Pizza equal twice the total number of pizza crusts made by Dianne's Delicatessen?
Mathematics
1 answer:
Paladinen [302]3 years ago
5 0
<span>Let pn  = The total number of pizza crusts </span>made by the end of the nth week

For Pepe's Pizza:
Pn + 52 + 180n 

For Diannes's 
Pn = 25 + 90n 

<span>The total number of pizza crusts made by Pep's Pizza equals twice the total number of pizza crusts made by Dianne's Delicatessen when:
52 + 180n = 2 (26 + 90n)
52 + 180n = 52 + 180n 

</span>This equation is satisfied for any value of n, so Pepe's always makes twice as much as Dianne's

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4 years ago
The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th
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97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

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:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

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

By the Central Limit Theorem

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

Z = \frac{30.9 - 29}{0.8341}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

X = 27.1

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

Z = \frac{27.1 - 29}{0.8341}

Z = -2.28

Z = -2.28 has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

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