If $\$ 1200$ are deposited into an account with $11 \%$ interest rate, compounded monthly, what is the balance after 11 years?
\[
\mathrm{F}=\$[?]
\]
Round to the nearest cent.
Enter
Answer
Finally, we need to check our answer. The balance after 11 years should be greater than the initial deposit of $1200, and our answer is indeed greater. Therefore, our answer is reasonable.
Step 1 ：First, we need to understand the problem. We are given an initial deposit of $1200, an annual interest rate of 11% compounded monthly, and we need to find the balance after 11 years.
Step 2 ：Next, we need to convert the annual interest rate to a monthly interest rate. Since there are 12 months in a year, the monthly interest rate is 11% divided by 12, or approximately 0.00916667.
Step 3 ：Then, we need to calculate the total number of periods. Since the interest is compounded monthly and we are looking at a period of 11 years, there are 11*12 = 132 periods.
Step 4 ：Now, we can use the formula for compound interest, which is \(P(1 + r/n)^{nt}\), where P is the principal amount (the initial amount of money), r is the annual interest rate (in decimal form), n is the number of times that interest is compounded per year, and t is the time the money is invested for in years.
Step 5 ：Substituting the given values into the formula, we get \(1200(1 + 0.00916667)^{132}\).
Step 6 ：Calculating the above expression, we get approximately $4024.46.
Step 7 ：So, the balance after 11 years is \(\boxed{4024.46}\).
Step 8 ：Finally, we need to check our answer. The balance after 11 years should be greater than the initial deposit of $1200, and our answer is indeed greater. Therefore, our answer is reasonable.
