To solve this problem, we use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n} \right)^{nt} \][/tex]
where:
- [tex]\(A\)[/tex] is the amount of money accumulated after [tex]\(n\)[/tex] years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial amount of money).
- [tex]\(r\)[/tex] is the annual interest rate (decimal).
- [tex]\(n\)[/tex] is the number of times that interest is compounded per year.
- [tex]\(t\)[/tex] is the number of years the money is invested or borrowed for.
Given the values:
- [tex]\( P = \$ 300 \)[/tex]
- [tex]\( r = 5\% = 0.05 \)[/tex]
- [tex]\( n = 4 \)[/tex] (since interest is compounded quarterly)
- [tex]\( t = 20 \)[/tex] years
We substitute these values into the formula:
[tex]\[ A = 300 \left( 1 + \frac{0.05}{4} \right)^{4 \times 20} \][/tex]
First, calculate the term inside the parentheses:
[tex]\[ 1 + \frac{0.05}{4} = 1 + 0.0125 = 1.0125 \][/tex]
Next, raise this term to the power of [tex]\( 4 \times 20 = 80 \)[/tex]:
[tex]\[ (1.0125)^{80} \][/tex]
Now, multiply this result by the principal amount [tex]\( P \)[/tex]:
[tex]\[ A = 300 \times (1.0125)^{80} \][/tex]
With all calculations done, you get:
[tex]\[ A \approx 810.45 \][/tex]
So, the investment will be worth approximately \[tex]$ 810.45 after 20 years. Among the given options, the correct one is:
\[ \$[/tex] 810.45 \]
