To find the account balance [tex]\( A \)[/tex] using the compound interest formula, we can use the following equation:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( P \)[/tex] is the principal amount, which is \[tex]$110,000.
- \( r \) is the annual interest rate (in decimal form). Given as 1.5%, we convert this to a decimal by dividing by 100, so \( r = \frac{1.5}{100} = 0.015 \).
- \( n \) is the number of times the interest is compounded per year. Since it is compounded annually, \( n = 1 \).
- \( t \) is the time the money is invested for, in years. Given \( t = 10 \) years.
Now, substituting the values into the compound interest formula:
\[ A = 110000 \left(1 + \frac{0.015}{1}\right)^{1 \cdot 10} \]
Simplify the inside of the parentheses first:
\[ 1 + \frac{0.015}{1} = 1.015 \]
Next, raise this number to the power of \((1 \cdot 10)\):
\[ (1.015)^{10} \]
Now, to get the final balance \( A \):
\[ A = 110000 \times (1.015)^{10} \approx 110000 \times 1.160538 \approx 127659.49\]
So, the account balance with the given conditions after 10 years is approximately:
\[ A \approx \$[/tex]127,659.49 \]
