To determine the balance in the account after 12 years with an initial deposit of [tex]$750, an annual interest rate of 9%, and quarterly compounding, we use the compound interest formula:
\[ F = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( P \) is the principal amount (initial deposit),
- \( r \) is the annual interest rate,
- \( n \) is the number of times interest is compounded per year,
- \( t \) is the number of years the money is invested.
Let's plug in the given values:
- \( P = 750 \)
- \( r = 0.09 \)
- \( n = 4 \)
- \( t = 12 \)
First, calculate the expression inside the parentheses:
\[ \frac{r}{n} = \frac{0.09}{4} = 0.0225 \]
Next, add 1 to this value:
\[ 1 + \frac{r}{n} = 1 + 0.0225 = 1.0225 \]
Raise this value to the power of \( n \times t \):
\[ n \times t = 4 \times 12 = 48 \]
\[ \left(1.0225\right)^{48} \]
Using the given result, we know that:
\[ \left(1.0225\right)^{48} \approx 2.909639612 \]
Finally, multiply this by the principal \( P = 750 \):
\[ F = 750 \times 2.909639612 \approx 2182.2297090673655 \]
Rounding to the nearest cent, the balance after 12 years is:
\[ F = 2182.23 \]
Thus, the balance after 12 years is \(\$[/tex]2182.23\).
