Question
George made an original investment of $1,000. The value of this investment increases by 5.5% each year. Which data point is represented on the graph for this situation?
A.
(1000, 0)
B.
(5, 1306.96)
C.
(1, 5500)
D.
(10, 2.53)
A.
(1000, 0)
B.
(5, 1306.96)
C.
(1, 5500)
D.
(10, 2.53)
Answers
GPT-4o mini
To determine which data point is represented on the graph for George's investment, we need to calculate the value of the investment after a certain number of years using the formula for compound interest:
\[
A = P(1 + r)^t
\]
where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the original investment).
- \(r\) is the annual interest rate (decimal).
- \(t\) is the time the money is invested for in years.
For this situation:
- \(P = 1000\)
- \(r = 0.055\) (5.5%)
- We will evaluate for \(t\) based on the options provided.
Let's calculate the values for the second option \(B\), which states (5, 1306.96).
Using \(t = 5\):
\[
A = 1000(1 + 0.055)^5
\]
Calculating \(1 + 0.055\):
\[
1 + 0.055 = 1.055
\]
Now calculating \(1.055^5\):
\[
1.055^5 \approx 1.303201
\]
Now multiply by 1000:
\[
A \approx 1000 \times 1.303201 \approx 1303.20
\]
The value after 5 years is approximately $1303.20, not the $1306.96 mentioned in option B. So, we continue checking other options.
Next, let's check option A:
Option A \( (1000, 0) \): This means at time \(t = 0\), the investment is \(1000\), which is accurate since it represents the initial investment.
Now for option C \( (1, 5500) \):
Calculating for \(t = 1\):
\[
A = 1000(1 + 0.055)^1 = 1000 \times 1.055 = 1055
\]
This does not match \(5500\).
Lastly, let's test option D \( (10, 2.53) \):
Calculating for \(t = 10\):
\[
A = 1000(1 + 0.055)^{10}
\]
Calculating \(1.055^{10}\):
\[
1.055^{10} \approx 1.7137
\]
Now multiply by 1000:
\[
A \approx 1000 \times 1.7137 \approx 1713.7
\]
This does not match \(2.53\).
Based on this, the only option that accurately represents the situation is:
**A. (1000, 0)**.
This is the point that represents the original investment at the time of investment.
\[
A = P(1 + r)^t
\]
where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the original investment).
- \(r\) is the annual interest rate (decimal).
- \(t\) is the time the money is invested for in years.
For this situation:
- \(P = 1000\)
- \(r = 0.055\) (5.5%)
- We will evaluate for \(t\) based on the options provided.
Let's calculate the values for the second option \(B\), which states (5, 1306.96).
Using \(t = 5\):
\[
A = 1000(1 + 0.055)^5
\]
Calculating \(1 + 0.055\):
\[
1 + 0.055 = 1.055
\]
Now calculating \(1.055^5\):
\[
1.055^5 \approx 1.303201
\]
Now multiply by 1000:
\[
A \approx 1000 \times 1.303201 \approx 1303.20
\]
The value after 5 years is approximately $1303.20, not the $1306.96 mentioned in option B. So, we continue checking other options.
Next, let's check option A:
Option A \( (1000, 0) \): This means at time \(t = 0\), the investment is \(1000\), which is accurate since it represents the initial investment.
Now for option C \( (1, 5500) \):
Calculating for \(t = 1\):
\[
A = 1000(1 + 0.055)^1 = 1000 \times 1.055 = 1055
\]
This does not match \(5500\).
Lastly, let's test option D \( (10, 2.53) \):
Calculating for \(t = 10\):
\[
A = 1000(1 + 0.055)^{10}
\]
Calculating \(1.055^{10}\):
\[
1.055^{10} \approx 1.7137
\]
Now multiply by 1000:
\[
A \approx 1000 \times 1.7137 \approx 1713.7
\]
This does not match \(2.53\).
Based on this, the only option that accurately represents the situation is:
**A. (1000, 0)**.
This is the point that represents the original investment at the time of investment.