To find the balance after 3 years in a savings account with an initial investment of \[tex]$1500 and a 4.6% annual compound interest rate, let's follow these steps:
1. Initial Investment: \$[/tex]1500.
2. Annual Interest Rate: 4.6%, which can be expressed as a decimal 0.046.
3. Number of Years: 3 years.
We use the compound interest formula:
[tex]\[ A = P(1 + r)^t \][/tex]
where
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal).
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Substitute the given values into the formula:
[tex]\[ A = 1500(1 + 0.046)^3 \][/tex]
First, calculate the growth factor:
[tex]\[ 1 + 0.046 = 1.046 \][/tex]
Next, raise this factor to the power of 3:
[tex]\[ 1.046^3 \approx 1.144445 \][/tex]
Now multiply this result by the initial investment:
[tex]\[ A \approx 1500 \times 1.144445 = 1716.668004 \][/tex]
Therefore, the balance after 3 years is approximately \$1716.668004.
Rounding to the nearest hundredth:
[tex]\[ 1716.668004 \approx 1716.67 \][/tex]
Thus, the balance after 3 years in the savings account will be:
[tex]\[ \boxed{1716.67} \][/tex]
