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tino4ka555 [31]
3 years ago
5

1. Peyton has a credit card with an annual rate of 24.7% compounded monthly. She used the credit card to purchase cleaning suppl

ies in the amount of $189.56. She can pay up to $72 on the
credit card each month. How much total interest will she pay?
Mathematics
1 answer:
enyata [817]3 years ago
4 0

Answer:

Total interest = $3.41

Step-by-step explanation:

Since she can pay $72 each month we can divide the payments on monthly basis till all the money is paid.

The annual interest rate is 24.7%, so the monthly rate will be 24.7 ÷ 12= 2.058%

For the first month

With payment of $72 the remaining amount will be 189.56 - 72 = $117.56

Interest paid will be 0.02058 * 117.56 = $2.42

Total amount owed now will be 117.56 + 2.42 = $119.98

For the second month another payment of $72 is made

The remaining will be 119.98 - 72 = $47.98

Interest charged will be 0.02058 * 47.98 = $0.99

The amount owed will be 47.98 + 0.99 = $48.97

In the third month she will pay the remaining $47.98 which is within her monthly limit

Total interest paid = Sum of Amount paid each month - Initial amount spent

Total interest = {(72 * 2) +48.97} - 189.56 = $3.41

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Step-by-step explanation:

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3 years ago
The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan
Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

\mu = E(X) = np = 100*0.23 = 23

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

So this is the pvalue of Z when X = 30.

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

Z = \frac{30 - 23}{4.1}

Z = 1.71

Z = 1.71 has a pvalue of 0.9564.

So there is a 95.64% probability that under 30% of the dogs are emotional support trained

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