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Callie entered an art contest in second grade and won a $1,000 scholarship. The money was invested in an account paying a 9% interest rate compounded annually. The situation can be modeled by the equation a(t)=1,000(1.09)t, where a is the amount in the account after t years. If Callie uses the scholarship 10 years later, determine which graph accurately displays the situation.(1 point)

Responses

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 5 in increments of 1 and the y-axis ranges from 0 to 2000 in increments of 500.
Image with alt text: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 5 in increments of 1 and the y-axis ranges from 0 to 2000 in increments of 500.

A decreasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 12 in increments of 2 and the y-axis ranges from 0 to 1,200 in increments of 200.
Image with alt text: A decreasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 12 in increments of 2 and the y-axis ranges from 0 to 1,200 in increments of 200.

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 10 in increments of 1 and the y-axis ranges from 0 to 2,000 in increments of 500.
Image with alt text: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 10 in increments of 1 and the y-axis ranges from 0 to 2,000 in increments of 500.

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 14 in increments of 2 and the y-axis ranges from 0 to 3,500 in increments of 500.

1 answer

To determine which graph accurately displays the situation of Callie's scholarship investment, we can analyze the growth of the investment over time using the given formula:

\[ a(t) = 1,000(1.09)^t \]

where \( t \) is the number of years.

Since Callie invested the money at a 9% annual interest rate compounded annually, the amount in the account grows exponentially over time.

Calculating the values after 10 years:

To find the value after 10 years:

\[ a(10) = 1,000(1.09)^{10} \]

Calculating \( (1.09)^{10} \):

Using a calculator, \[ (1.09)^{10} \approx 2.36736 \]

Then:

\[ a(10) = 1,000 \times 2.36736 \approx 2367.36 \]

This means that after 10 years, the investment will be approximately $2,367.36.

Analyzing Graph Options:

Now we need to look for a graph that represents an increasing curve (since the amount is increasing due to compound interest), ideally showing values that reach around $2,367 after 10 years.

  1. First Option: X-axis from 0 to 5 (shows less than 10 years)
  2. Second Option: A decreasing curve (not applicable as the investment is increasing)
  3. Third Option: X-axis from 0 to 10 and Y-axis ranging up to 2,000 (could be below $2,367)
  4. Fourth Option: X-axis from 0 to 14 (this one may fit, depending on how far it goes above 2,000)

Given that at 10 years, we calculated around $2,367, the best option would be:

  • Third Option: An increasing curve that spans from 0 to 10 with a maximum of 2,000. This graph is likely a better representation of the investment over time as it focuses specifically on the first 10 years of growth, suggesting a value close to $2,367.

Thus, the third option accurately displays the situation, as the investment continues to increase, at least indicating the general behavior of the investment over that time frame, even if it doesn't reach the final calculated value.