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Vesnalui [34]
3 years ago
7

Eddie took out a 14-year loan for $72,000 at an APR of 4.7%, compounded monthly, while Lee took out a 14-year loan for $92,000 a

t an APR of 4.7%, compounded monthly. Who would save more by paying off his loan 6 years early?
A. Lee would save more, since he has $20,000 more in principal.
B. Lee would save more, since he has $20,000 less in principal.
C. Eddie would save more, since he has $20,000 less in principal.
D. Eddie would save more, since he has $20,000 more in principal.
Mathematics
2 answers:
boyakko [2]3 years ago
7 0

Answer:

Lee

Step-by-step explanation:

It is given in the problem that Lee has taken a loan of $ 92,000 while Eddie has taken a loan of only $72,000.

however, the rate of interest and the tenure is the same for both of them. Hence, it is the principle which is going to affect the interest incurred on Lee or Eddie.

As Lee has higher amount of loan , the interest ion him will be high as compared to the Eddie. Thus Lee will save more if both of them pay their debt in 8 years that is 6 years before the due time.

marishachu [46]3 years ago
4 0
**Eddie: $72000/(14yr*12mo)=428.6$/mo+428.6$*(4.7%)/100%
Eddie pays 428.6$/mo+20.14$/mo. If he pays off his loan 6 years earlier he would save: $20.14*6yr*12mo= $1450.08
**Lee: $92000/(14yr*12mo)=547.62$/mo+547.62$*(4.7%)/100%
Lee pays 547.62$/mo+25.74$/mo. If he pays off his loan 6 years earlier he would save: $25.74*6yr*12mo=$1853.28

So its A. <span>Lee would save more, since he has $20,000 more in principal.</span>
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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a
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Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

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Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

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The standard deviation of the the sampling distribution of the sample proportion is:

SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092

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Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: p=\frac{5}{25} =0.20

P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: p=\frac{7}{25} =0.28

P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118

The exact probability is more than the approximated probability.

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Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: p=\frac{9}{25} =0.36

P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106

The exact probability is more than the approximated probability.

(d)

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The probability of 5 successes is: p=\frac{11}{25} =0.44

P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357

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P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558

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