The question from your screenshot asks:

"How much money should be deposited today in an account that earns 4.5% compounded monthly so that it will accumulate to $15,000 in 2 years?"

This is a typical problem involving compound interest, and we can solve it using the following compound interest formula:

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

Where:

• $A$ is the future value of the investment/loan (the amount you want to accumulate), which is $15,000.
• $P$ is the principal amount (the money you need to deposit today).
• $r$ is the annual interest rate (4.5%, or 0.045).
• $n$ is the number of times the interest is compounded per year (since it's compounded monthly, $n = 12$).
• $t$ is the time the money is invested or borrowed for, in years (in this case, $t = 2$).

We want to find $P$, the amount to be deposited today. Rearranging the formula to solve for $P$:

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

Now, we can plug in the values:

$P = \frac{15,000}{\left(1 + \frac{0.045}{12}\right)^{12 \times 2}}$

Let me calculate this for you.The amount of money that should be deposited today is approximately $13,711.28.

Would you like further clarification on how this was calculated, or any additional details?

Here are some related questions for further exploration:

1. How does the compounding frequency affect the amount you need to deposit?
2. What if the interest rate was higher or lower? How would that change the result?
3. How would the deposit amount change if the time period were longer, such as 5 years instead of 2 years?
4. What would happen if the interest were compounded quarterly instead of monthly?
5. How does continuous compounding differ from monthly compounding?

Tip: Always make sure to convert percentage interest rates to decimal form (e.g., 4.5% becomes 0.045) when using formulas.