The situation describes an investment growing due to compound interest. The equation given is:
\[ a(t) = 1,000(1.09)^t \]
This equation represents exponential growth, where the amount \( a(t) \) increases as time \( t \) (in years) increases.
Since Callie is using the scholarship 10 years later, we need to focus on the behavior of the function within that time frame, specifically from \( t = 0 \) to \( t = 10 \).
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Increasing Curve: The function \( a(t) \) is increasing because the principal (1,000) and the growth factor (1.09) are both positive, and exponential functions grow as time increases.
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Relevant Range: For \( t \) values, we are interested in the range from 0 to 10 years.
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Amount Range: After 10 years, the amount in the account would be:
\[ a(10) = 1,000(1.09)^{10} \]
Calculation of \( (1.09)^{10} \): \[ (1.09)^{10} \approx 2.367364 \] Thus: \[ a(10) \approx 1,000 \times 2.367364 \approx 2367.36 \]
Given these points, we can determine which graph fits this scenario:
- It must be an increasing curve.
- The x-axis should extend to at least 10.
- The y-axis should display amounts around 2367.
Looking through your four responses:
- The first option shows a decreasing curve, which doesn’t fit.
- The second option is an increasing curve from 0 to 10 with the y-axis reaching up to 2,000. It may not show the maximum amount of around 2367 accurately.
- The third option has an increasing curve but only goes up to 5 years; it doesn't cover the full range.
- The fourth option has an increasing curve that goes beyond 10 years and the y-axis goes up to a maximum of 3,500, which can accommodate the calculated amount.
Based on the criteria, the best option is the fourth response, as it shows an increasing exponential function, covers the necessary time frame, and can encapsulate the calculated value after 10 years.