Question

Mariana is going to invest $6,200 and leave it in an account for 9 years. Assuming the

interest is compounded monthly, what interest rate, to the nearest tenth of a percent,
would be required in order for Mariana to end up with $7,400?

Answer 1

102.15% per year more

Answer 2

The interest rate required for Mariana to end up with $7,400 after 9 years with an initial investment of $6,200, and assuming the interest is compounded monthly, would be approximately 2.8%.

To find the interest rate required for Mariana to end up with $7,400 after 9 years with monthly compounding, we can use the formula for compound interest:

`\[ A = P \left(1 + (r)/(n)\right)^(nt) \]`

where:

- A is the future value of the investment (in this case, $7,400),

- P is the principal amount (initial investment, $6,200),

- r is the annual interest rate (what we're trying to find),

- n is the number of times interest is compounded per year (monthly compounding means n = 12 ,

- t is the time the money is invested or borrowed for, in years (9 years in this case).

The formula for compound interest is given by:

`\[ A = P \left(1 + (r)/(n)\right)^(nt) \]`

In this scenario, Mariana wants to end up with $7,400: A = 7400, starts with an initial investment of $6,200 : P = 6200, invests for 9 years t = 9, and interest is compounded monthly n = 12.

We rearrange the formula to solve for r:

`\[ r = n \left(\left((A)/(P)\right)^{(1)/(nt)} - 1\right) \]`

Substitute the known values:

`\[ r = 12 \left(\left((7400)/(6200)\right)^{(1)/(12 * 9)} - 1\right) \]`

Calculate the expression:

`\[ r \approx 0.028 \]`

Convert to a percentage:

`\[ r \approx 0.028 * 100\% \approx 2.8\% \]`

Therefore, Mariana would need an interest rate of approximately 2.8% (to the nearest tenth of a percent) for her investment to grow to $7,400 after 9 years with monthly compounding.