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Brums [2.3K]
3 years ago
10

Jason makes a credit card purchase for $3000 that compounds interest daily. The APR is 4%. How much interest is earned on Jason'

s card after 1 year if Jason makes no payments?
The amount of interest earned on Jason's card after 1 year is $
Mathematics
1 answer:
Vlad [161]3 years ago
6 0

Step-by-step explanation:

step 1. A = P(1 + r/n)^nt. this is the compounding equation where n is the number of compounds in a time t and P is the original amount. r is the rate.

step 2. A = 3000(1 + .04/365)^(365(1))

step 3. A = 3122.43. this is the total amount owed on the credit card which includes principle and interest. the interest is less 5he principle

step 4. 3122.43 - 3000 = 122.43

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a) Colors of phone cover - quantitative

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A company developing a new cellular phone plan intends to market their new phone to customers who use text and social media ofte
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Answer:

A customer who sends 78 messages per day would be at 99.38th percentile.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Average of 48 texts per day with a standard deviation of 12.

This means that \mu = 48, \sigma = 12

a. A customer who sends 78 messages per day would correspond to what percentile?

The percentile is the p-value of Z when X = 78. So

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

Z = \frac{78 - 48}{12}

Z = 2.5

Z = 2.5 has a p-value of 0.9938.

0.9938*100% = 99.38%.

A customer who sends 78 messages per day would be at 99.38th percentile.

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