To analyze the function \( f(t) = 4,000(1.0056)^{12t} \):
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The initial amount of money invested is represented by the coefficient of the function when \( t = 0 \). Plugging in \( t = 0 \): \[ f(0) = 4,000(1.0056)^{12 \cdot 0} = 4,000(1) = 4,000 \] Therefore, the initial amount of money invested is $4,000.
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The function represents an exponential model. The growth factor inside the parentheses, \( 1.0056 \), can be converted into a percentage to determine the annual growth rate. Since the amount is compounded monthly, we'll first find the effective annual rate.
The effective annual rate (EAR) with monthly compounding can be calculated using the formula: \[ \text{EAR} = (1 + \text{monthly rate})^{12} - 1 \] In this case, the monthly rate is \( 0.0056 \). Thus: \[ \text{EAR} = (1 + 0.0056)^{12} - 1 \approx 1.0698 - 1 \approx 0.0698 \] Converting this to a percentage: \[ 0.0698 \times 100 \approx 6.98% \]
Therefore, the annual percentage increase in the amount of money in the account is approximately 6.98%.
Summary:
- The initial amount of money invested is $4,000.
- The function represents an exponential model where the amount of money in the account increases by approximately 6.98% each year.