To determine how much Jeff had in his account after 3 years, we will use the formula for compound interest, which is given by:
[tex]\[ A(t) = P \cdot (1 + i)^t \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money)
- [tex]\( i \)[/tex] is the annual interest rate (expressed as a decimal)
- [tex]\( t \)[/tex] is the number of years the money is invested or borrowed for
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
Given values:
- Principal, [tex]\( P = 3000 \)[/tex]
- Annual interest rate, [tex]\( i = 6.5\% = 0.065 \)[/tex]
- Number of years, [tex]\( t = 3 \)[/tex]
Substituting these values into the formula, we get:
[tex]\[ A(3) = 3000 \cdot (1 + 0.065)^3 \][/tex]
First, calculate the term inside the parentheses:
[tex]\[ 1 + 0.065 = 1.065 \][/tex]
Now, raise this to the power of 3:
[tex]\[ 1.065^3 \approx 1.207728 \][/tex]
Multiply this result by the principal amount:
[tex]\[ 3000 \cdot 1.207728 \approx 3623.85 \][/tex]
Therefore, after 3 years, Jeff had approximately \[tex]$ 3623.85 in the account.
The correct answer is:
D. \$[/tex] 3623.85
