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Mr. Bert deposited $5,000 into an investment account 20 years ago with an annual interest rate of 3.75%. a) Find the exponentia...Question
mr bert deposited 5,000 into an investment account 20 years ago with an annual interest rate of 3.75 percent
a find the exponential function in the dorm f(x)=atimesb^x to represent the total value od mrsberts investment account over time
b mr bert will retire in 15 years what will be the total value of his investments when he retires round the answer to 2 decimals
c what is the percent increase of this final amount compared to the initial amount of the investment
a find the exponential function in the dorm f(x)=atimesb^x to represent the total value od mrsberts investment account over time
b mr bert will retire in 15 years what will be the total value of his investments when he retires round the answer to 2 decimals
c what is the percent increase of this final amount compared to the initial amount of the investment
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Answered by
GPT-4o mini
To solve the problem, we will follow the steps outlined in parts a, b, and c.
### Part a: Finding the Exponential Function
The formula for exponential growth in the context of compound interest is given by:
\[
A(t) = A_0 \times (1 + r)^t
\]
where:
- \( A(t) \) = the amount of money accumulated after n years, including interest.
- \( A_0 \) = the principal amount (the initial deposit or investment).
- \( r \) = annual interest rate (in decimal).
- \( t \) = time in years.
In this case:
- \( A_0 = 5000 \)
- \( r = 3.75\% = 0.0375 \)
Therefore, we can express the function as:
\[
A(t) = 5000 \times (1 + 0.0375)^t
\]
Simplifying this gives:
\[
A(t) = 5000 \times (1.0375)^t
\]
This fits the form \( f(x) = a \times b^x \), where \( a = 5000 \) and \( b = 1.0375 \). Thus,
\[
f(t) = 5000 \times (1.0375)^t
\]
### Part b: Total Value of Investments in 15 Years
Mr. Bert will retire in 15 years from now. Since his investment has already been in place for 20 years, we need to calculate the total duration of investment at retirement, which will be \( 20 + 15 = 35 \) years.
Now we calculate:
\[
A(35) = 5000 \times (1.0375)^{35}
\]
Calculating \( (1.0375)^{35} \):
\[
(1.0375)^{35} \approx 3.4637
\]
So,
\[
A(35) \approx 5000 \times 3.4637 \approx 17318.50
\]
Thus, the total value of Mr. Bert's investments at retirement is approximately **$17,318.50**.
### Part c: Percent Increase Compared to the Initial Amount
The percent increase is calculated using the formula:
\[
\text{Percent Increase} = \frac{\text{Final Amount} - \text{Initial Amount}}{\text{Initial Amount}} \times 100
\]
Substituting in the initial amount (\$5000) and final amount (\$17318.50):
\[
\text{Percent Increase} = \frac{17318.50 - 5000}{5000} \times 100
\]
Calculating this gives:
\[
\text{Percent Increase} = \frac{12318.50}{5000} \times 100 \approx 246.37\%
\]
### Summary of Results:
- **a**. The exponential function is \( f(t) = 5000 \times (1.0375)^t \).
- **b**. The total value of Mr. Bert's investments when he retires will be approximately **$17,318.50**.
- **c**. The percent increase compared to the initial amount of the investment is approximately **246.37%**.
### Part a: Finding the Exponential Function
The formula for exponential growth in the context of compound interest is given by:
\[
A(t) = A_0 \times (1 + r)^t
\]
where:
- \( A(t) \) = the amount of money accumulated after n years, including interest.
- \( A_0 \) = the principal amount (the initial deposit or investment).
- \( r \) = annual interest rate (in decimal).
- \( t \) = time in years.
In this case:
- \( A_0 = 5000 \)
- \( r = 3.75\% = 0.0375 \)
Therefore, we can express the function as:
\[
A(t) = 5000 \times (1 + 0.0375)^t
\]
Simplifying this gives:
\[
A(t) = 5000 \times (1.0375)^t
\]
This fits the form \( f(x) = a \times b^x \), where \( a = 5000 \) and \( b = 1.0375 \). Thus,
\[
f(t) = 5000 \times (1.0375)^t
\]
### Part b: Total Value of Investments in 15 Years
Mr. Bert will retire in 15 years from now. Since his investment has already been in place for 20 years, we need to calculate the total duration of investment at retirement, which will be \( 20 + 15 = 35 \) years.
Now we calculate:
\[
A(35) = 5000 \times (1.0375)^{35}
\]
Calculating \( (1.0375)^{35} \):
\[
(1.0375)^{35} \approx 3.4637
\]
So,
\[
A(35) \approx 5000 \times 3.4637 \approx 17318.50
\]
Thus, the total value of Mr. Bert's investments at retirement is approximately **$17,318.50**.
### Part c: Percent Increase Compared to the Initial Amount
The percent increase is calculated using the formula:
\[
\text{Percent Increase} = \frac{\text{Final Amount} - \text{Initial Amount}}{\text{Initial Amount}} \times 100
\]
Substituting in the initial amount (\$5000) and final amount (\$17318.50):
\[
\text{Percent Increase} = \frac{17318.50 - 5000}{5000} \times 100
\]
Calculating this gives:
\[
\text{Percent Increase} = \frac{12318.50}{5000} \times 100 \approx 246.37\%
\]
### Summary of Results:
- **a**. The exponential function is \( f(t) = 5000 \times (1.0375)^t \).
- **b**. The total value of Mr. Bert's investments when he retires will be approximately **$17,318.50**.
- **c**. The percent increase compared to the initial amount of the investment is approximately **246.37%**.
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