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The problem in the image asks how much money should be deposited today in an account that earns 6.5% interest compounded monthly, in order to accumulate to $12,000 in three years.

This is a typical future value problem, where we want to find the present value. The formula for compound interest is:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $A$ is the future value, which is $12,000.
- $P$ is the present value, the amount to be deposited today (what we need to find).
- $r$ is the annual interest rate, which is 6.5% or 0.065.
- $n$ is the number of times the interest is compounded per year (monthly compounding means $n = 12$).
- $t$ is the number of years, which is 3.

We need to rearrange the formula to solve for $P$:

\[
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
\]

Now let's plug in the values and calculate. I will do this calculation now.The amount of money that should be deposited today is approximately **$9,879.21**.

Would you like further details or have any questions?

Here are five related questions for further exploration:
1. How would the deposit amount change if the interest were compounded annually instead of monthly?
2. What would be the deposit amount if the goal were $15,000 in three years at the same interest rate?
3. How much interest would be earned in the three years on this deposit?
4. What would be the deposit amount if the interest rate were 5% instead of 6.5%?
5. How does the frequency of compounding affect the required deposit?

**Tip:** Compounding frequency plays a crucial role in determining the future value of an investment—higher frequency generally leads to more interest earned over time.

How much money should be deposited today in an account that earns 6.5% compounded monthly so that it will accumulate to $12,000 in three years?

Find out how much money needs to be deposited today to accumulate $12,000 in three years, with a 6.5% interest rate compounded monthly. Learn to use the compound interest formula for present value and future value calculations. Suitable for high school students in Grades 11-12.

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