To find the balance in the account after 3 years for a deposit of $5000 earning 1.5% interest compounded quarterly, we can use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial deposit or investment).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for in years.
In this case:
- \( P = 5000 \)
- \( r = 0.015 \) (1.5% as a decimal)
- \( n = 4 \) (quarterly)
- \( t = 3 \)
Plugging in these values:
\[ A = 5000 \left(1 + \frac{0.015}{4}\right)^{4 \times 3} \]
Calculating the interest rate per compounding period:
\[ \frac{0.015}{4} = 0.00375 \]
Now substituting this into the formula:
\[ A = 5000 \left(1 + 0.00375\right)^{12} \]
Calculating inside the parentheses:
\[ 1 + 0.00375 = 1.00375 \]
Now, raise this to the power of 12:
\[ A = 5000 \left(1.00375^{12}\right) \]
Calculating \(1.00375^{12}\):
\[ 1.00375^{12} \approx 1.045682 (can vary slightly based on calculator precision) \]
Now multiplying by the principal:
\[ A = 5000 \times 1.045682 \approx 5228.41 \]
So, the balance in the account after 3 years is approximately:
\[ \boxed{5228.41} \]
Looking at the provided options, the closest answer to our calculation of approximately $5228.41 is $5,229.70.
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