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garik1379 [7]
3 years ago
12

Gavin has $7,500 to invest. He is considering two investment options. Option A pays 4% simple interest. Option B pays 3.15% inte

rest compounded annually. Drag dollar amounts to the table to show the value of each investment option after 4 years and after 6 years rounded to the nearest dollar.
Mathematics
2 answers:
ElenaW [278]3 years ago
7 0

Answer:

4 years Option A: $8,700

6 years Option B: $9034

Leviafan [203]3 years ago
4 0

Solution:

Principal =P= $ 7,500

Option A→(Simple interest)

Rate of interest= R=4%

Time(T_{1})=4 years

Time(T_{2})=6 years

Amount= Principal + Interest(Simple or compound interest)

Formula for Simple interest

S.I=\frac{P\times R\times T}{100}

S.I_{1}=\frac{7500 *4*4}{100}=1200\\\\ S.I_{2}=\frac{7500 *4*6}{100}=1800

Total amount after 4 years when interest is simple= 7500 +1200= $ 8700

Total amount after 6 years  when interest is simple= 7500 +1800= $ 9300

Option B

Formula for amount(A) when interest is 3.15% compounded annually.

A=P*(1+\frac{R}{100})^t

A_{4}=7500*(1+\frac{3.15}{100})^4\\\\ A_{4}=7500*(\frac{103.15}{100})^4\\\\ A_{4}=7500*(1.0315)^4\\\\ A_{4}=7500*1.1320\\\\ A_{4}=8490.60

A_{6}=7500*(1+\frac{3.15}{100})^6\\\\ A_{6}=7500*(\frac{103.15}{100})^6\\\\ A_{6}=7500*(1.0315)^6\\\\ A_{6}=7500*1.2045\\\\ A_{6}=9033.9286

Total amount after 4 years when interest is compounded annually=$ 8491 (approx)

Total amount after 6 years  when interest is compounded annually=$ 9034(approx)

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Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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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}

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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 = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

Z = \frac{21 - 18.6}{5.4}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

By the Central Limit Theorem

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

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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