To find Gerry's balance at the close of December 31, 2012, we first determine how many years the savings account was open. Since he opened the account on January 1, 2003, and was considering the balance at the close of December 31, 2012, the account was open for:
- From January 1, 2003, to December 31, 2012: 2012 - 2003 + 1 = 10 years.
Next, we use the formula for compound interest to calculate the balance. The formula is:
\[ A = P(1 + r)^t \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial deposit).
- \( r \) is the annual interest rate (decimal).
- \( t \) is the number of years the money is invested or borrowed.
In Gerry's case:
- \( P = 1100 \) (the initial deposit),
- \( r = 3.25% = 0.0325 \) (the annual interest rate in decimal form),
- \( t = 10 \) (the number of years).
Substituting these values into the formula:
\[ A = 1100(1 + 0.0325)^{10} \]
Calculating \( 1 + 0.0325 \):
\[ 1 + 0.0325 = 1.0325 \]
Now calculating \( (1.0325)^{10} \):
\[ (1.0325)^{10} \approx 1.343916 \]
Now, substituting this back into the formula:
\[ A \approx 1100 \times 1.343916 \approx 1488.3086 \]
Rounding this to two decimal places, Gerry's balance at the close of December 31, 2012, would be approximately:
\[ \boxed{1488.31} \]
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