To solve this problem, we can use the formula for compound interest, which is:

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

Where:

• $A$ is the amount of money accumulated after $t$ years, including interest ($10,000 in this case).
• $P$ is the principal amount (the initial deposit, which we need to find).
• $r$ is the annual interest rate (3.5% = 0.035).
• $n$ is the number of times the interest is compounded per year (monthly, so $n = 12$).
• $t$ is the time the money is invested for, in years (3 years).

Step 1: Plug in the known values

• $A = 10,000$
• $r = 0.035$
• $n = 12$
• $t = 3$

The formula becomes:

$10,000 = P \left( 1 + \frac{0.035}{12} \right)^{12 \times 3}$

Step 2: Simplify the equation

First, calculate the monthly interest rate:

$\frac{0.035}{12} = 0.0029167$

Now calculate $12 \times 3 = 36$, which is the total number of months.

$10,000 = P \left( 1 + 0.0029167 \right)^{36}$

$10,000 = P \left( 1.0029167 \right)^{36}$

Now, calculate $\left( 1.0029167 \right)^{36}$:

$1.0029167^{36} \approx 1.1138$

Step 3: Solve for $P$

Now the equation looks like this:

$10,000 = P \times 1.1138$

To find $P$, divide both sides by 1.1138:

$P = \frac{10,000}{1.1138} \approx 8,978.64$

Final Answer:

The amount that should be deposited today is $8,978.64 (rounded to the nearest cent).

Let me know if you'd like more details on any part of the process or have any questions!

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Related Questions:

1. How does compound interest differ from simple interest?
2. What is the effect of increasing the frequency of compounding (e.g., quarterly or daily)?
3. How would the deposit amount change if the interest rate was 4% instead of 3.5%?
4. What happens to the final accumulated amount if the investment period is extended to 5 years?
5. How do you calculate the monthly interest rate for different compounding periods?

Tip: Always ensure to convert the interest rate to a decimal before plugging it into the formula (e.g., 3.5% becomes 0.035).