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Schach [20]
2 years ago
13

A loan of $900 will be repaid in 36 installment payments of $30.22 each. What the finance charge for the loan?

Mathematics
1 answer:
zimovet [89]2 years ago
3 0
First you need to multiply 36 *$30.22=$1087.92
This gives you the total amount paid

Then subtract 1087.92-900=187.92
This gives you the amount of money you paid in finance charges.
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The graph below represents the distance a pilot is from the home airport during a period of time. A graph titled Pilot's Distanc
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Step-by-step explanation:

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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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Answer:

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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2 years ago
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3. Atile warehouse has Inventory at hand and can put in for a
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Step-by-step explanation:

Given the equation that represents this order expressed as;

The number of tiles = 12b + 38 where;

b is the the number of bundles ordered

If a customer needs 150 tiles, the total number of bundles ordered can be gotten by simply substituting The number of tiles into the modeled equation and find the value of b. This is as shown below;

On substituting;

150 = 12b + 38

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We need to round up the problem because the number of tiles can not be in fraction but as whole numbers.

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This means we can draw a line of best fit through this data and it should hit several data points; thus the answer is linear.

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