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Compound Interest An investment of $5000 is deposited into an account in which interest is compounded monthly. Complete the table by filling in the amounts to which the investment grows at the indicated times or interest rates.

r=4%

Short Answer

Expert verified

Time (years)

Amount

1

5203.70

2

5415.71

3

5636.36

4

5865.99

5

6104.98

6

6353.71

Step by step solution

01

Step 1. Given.

P=Principal=5000.

r=interestrate=4%=0.04.

n=The number of times compounded in a year =12.

Because, interest is compounded monthly.

02

Step 2. To determine.

We have to complete the table by filling in the amounts to which the investment grows at the indicated times.

03

Step 3. Calculation.

Given:

P=Principal=5000.

r=interestrate=4%=0.04.

n=The number of times compounded in a year =12.

t=number of years.

A(t)=Amount after t years

Here,

A(t)=P1+rnnt=50001+0.041212t

We have to find A(t) for t=1,2,3,4,5,6. We plug these values in A(t).

A(1)=50001+0.041212(1)=5203.70771465203.71

A(2)=50001+0.041212(2)=5415.71479585415.71

A(3)=50001+0.041212(3)=5636.359372595636.36

A(4)=50001+0.041212(4)=5865.993349885865.99

A(5)=50001+0.041212(5)=6104.982969716104.98

A(6)=50001+0.041212(6)=6353.70939546353.71

Hence the table is:

Time (years)

Amount

1

5203.70

2

5415.71

3

5636.36

4

5865.99

5

6104.98

6

6353.71

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