To determine how much money should be deposited today to accumulate $11,000 in 4 years with an interest rate of 4.5% compounded monthly, we can use the formula for compound interest:

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

Where:

• $A$ is the future value (what we want to accumulate, $11,000).
• $P$ is the present value (the initial deposit, which we're solving for).
• $r$ is the annual interest rate (4.5%, or 0.045).
• $n$ is the number of times the interest is compounded per year (monthly compounding means $n = 12$).
• $t$ is the time in years (4 years).

We need to solve for $P$. Rearranging the formula to solve for $P$:

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

Now, plugging in the values:

$P = \frac{11,000}{\left(1 + \frac{0.045}{12}\right)^{12 \times 4}}$

Let's calculate this:

$P = \frac{11,000}{\left(1 + 0.00375\right)^{48}}$

$P = \frac{11,000}{\left(1.00375\right)^{48}}$

$P = \frac{11,000}{1.197992}$

$P \approx 9,181.49$

So, you would need to deposit approximately $9,181.49 today to accumulate $11,000 in 4 years at an interest rate of 4.5%, compounded monthly.

Do you want more details or have any questions about this?

Here are 5 related questions:

1. How would the answer change if the interest were compounded annually instead of monthly?
2. What happens to the required initial deposit if the interest rate increases to 5%?
3. How would the formula adjust for a quarterly compounding period?
4. How much interest would be earned after 4 years on the initial deposit?
5. What if the investment goal was $15,000 instead of $11,000?

Tip: When dealing with compound interest, the more frequently the interest is compounded, the greater the accumulated amount will be.