The problem you've uploaded involves calculating the initial deposit in a savings account, given the interest earned and the balance after a certain period. This can be solved using the formula for compound interest:

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

Where:

• $A$ is the final amount (current balance), which is $500.00.
• $P$ is the initial deposit (the amount we need to find).
• $r$ is the annual interest rate (4% or 0.04).
• $n$ is the number of times the interest is compounded per year (monthly, so $n = 12$).
• $t$ is the time in years (2 years).

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

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

Substituting the known values:

$P = \frac{500}{\left( 1 + \frac{0.04}{12} \right)^{12 \times 2}}$

Now, let's compute this step by step:

1. $\frac{0.04}{12} = 0.0033333$
2. $1 + 0.0033333 = 1.0033333$
3. $12 \times 2 = 24$
4. $1.0033333^{24} \approx 1.083282$
5. $P = \frac{500}{1.083282} \approx 461.51$

So, the initial deposit was approximately $461.51.

Would you like further details on any steps or clarifications?

Here are 5 related questions for further exploration:

1. How would the initial deposit change if the interest rate was higher?
2. What would the balance be after 5 years with the same initial deposit and interest rate?
3. How does the compounding frequency affect the final balance in this scenario?
4. What would happen if the interest were compounded annually instead of monthly?
5. How can you use this formula to calculate the interest earned over a period?

Tip: When dealing with compound interest, the frequency of compounding has a significant effect on the amount accumulated over time. More frequent compounding results in a higher balance.