If a nurse deposits $12,000 today in a bank account and the interest is compounding annually at 11%, what will be the value of this investment nine years from now?
The future value of the money would be the original 12000 plus 11%, or 12000 times 1.11, after one year. After 9 years, the money will be worth 12000 times (1.11)^9, or 30696.44
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To find the value of the investment after nine years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
- A is the final amount after the given time period
- P is the principal amount (the initial deposit)
- r is the annual interest rate (in decimal form)
- n is the number of times the interest is compounded per year
- t is the number of years
In this case, the nurse deposited $12,000, the interest rate is 11% (0.11 in decimal form), and the interest is compounded annually (n = 1) for 9 years (t = 9).
Plugging these values into the formula, we have:
A = 12000(1 + 0.11/1)^(1*9)
Simplifying:
A = 12000(1 + 0.11)^9
Now, we can calculate the value of the investment by raising (1 + 0.11) to the power of 9, and then multiply by $12,000:
A = 12000(1.11)^9
Using a calculator, we can find the value of (1.11)^9 is approximately 2.3773.
A = 12000 * 2.3773
Calculating:
A ≈ $28,527.60
Therefore, the value of this investment after nine years will be approximately $28,527.60.