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Part 1
Find the value of an investment of $ 10 comma 000 for 11 years at an annual interest rate of 4.55 %     compounded continuously.
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Part 1
The value of the investment is ​$

enter your response here.
​(Do not round until the final answer. Then round to the nearest cent as​ needed.)

To solve this, we will use the formula for continuous compounding:

\[
A = Pe^{rt}
\]

where:
- $ A $ is the amount of money accumulated after $ t $ years, including interest,
- $ P $ is the principal investment amount (the initial deposit or loan amount),
- $ r $ is the annual interest rate (decimal),
- $ t $ is the number of years the money is invested for, and
- $ e $ is the base of the natural logarithm (approximately 2.71828).

### Given values:
- $ P = 10,000 $
- $ r = 0.0455 $
- $ t = 11 $

### Step-by-Step Solution
Substitute the values into the formula:

\[
A = 10,000 \times e^{(0.0455 \times 11)}
\]

Calculating $ 0.0455 \times 11 $:

\[
0.0455 \times 11 = 0.5005
\]

So,

\[
A = 10,000 \times e^{0.5005}
\]

Using $ e^{0.5005} \approx 1.6499 $ (rounded to four decimal places for accuracy):

\[
A = 10,000 \times 1.6499
\]

Now, multiplying:

\[
A \approx 16,499
\]

### Final Answer
The value of the investment is approximately:

\[
A \approx 16,499.00
\]

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Would you like further clarification on any step?

### Related Questions
1. What if the interest rate was compounded annually instead of continuously?
2. How would the answer change if the investment period was shorter, say 5 years?
3. How can the formula be used for monthly compounding?
4. What happens if the principal is doubled?
5. How is the continuous compounding formula derived?

### Tip
Always make sure to convert percentages into decimals before substituting into formulas for accurate results.

Find the value of an investment of $10,000 for 11 years at an annual interest rate of 4.55% compounded continuously.

Learn to calculate the future value of a $10,000 investment over 11 years with continuous compounding at an annual rate of 4.55%. This guide covers exponential growth principles, continuous compounding formula, and interest calculations with step-by-step solutions.

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