An investment of $5,000 is deposited into an account for which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the indicated times. (Round your answers to the nearest cent.)
r = 6%
Time (years)	Amount
1	$ 
5309.22
 
Incorrect: Your answer is incorrect.
2	$ 
5637.50
 
Incorrect: Your answer is incorrect.

See the rounding prompt for how many decimal places are needed.
3	$ 
5970.25An investment of $5,000 is deposited into an account for which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the indicated times. (Round your answers to the nearest cent.)
r = 6%
Time (years)	Amount
1	$ 
5309.22
 
Incorrect: Your answer is incorrect.
2	$ 
5637.50
 
Incorrect: Your answer is incorrect.

See the rounding prompt for how many decimal places are needed.
3	$ 
5970.25
 
Incorrect: Your answer is incorrect.
4	$ 
6309.55
 
Incorrect: Your answer is incorrect.
5	$ 
6657.00
 
Incorrect: Your answer is incorrect.
6
 
Incorrect: Your answer is incorrect.
4	$ 
6309.55
 
Incorrect: Your answer is incorrect.
5	$ 
6657.00
 
Incorrect: Your answer is incorrect.
6

To solve this problem, we need to use the formula for **continuously compounded interest**:

\[
A = P e^{rt}
\]

Where:
- $ A $ is the amount after time $ t $,
- $ P $ is the initial principal (in this case, $5000),
- $ r $ is the interest rate (in this case, 6%, or 0.06 as a decimal),
- $ t $ is the time in years, and
- $ e $ is the base of the natural logarithm (approximately 2.71828).

Let's go through each year:

### For $ t = 1 $ year:
\[
A = 5000 \times e^{0.06 \times 1}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.06} \approx 5000 \times 1.06183655 \approx 5309.18
\]

### For $ t = 2 $ years:
\[
A = 5000 \times e^{0.06 \times 2}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.12} \approx 5000 \times 1.12749685 \approx 5637.48
\]

### For $ t = 3 $ years:
\[
A = 5000 \times e^{0.06 \times 3}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.18} \approx 5000 \times 1.19721736 \approx 5986.09
\]

### For $ t = 4 $ years:
\[
A = 5000 \times e^{0.06 \times 4}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.24} \approx 5000 \times 1.27124915 \approx 6356.25
\]

### For $ t = 5 $ years:
\[
A = 5000 \times e^{0.06 \times 5}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.30} \approx 5000 \times 1.34985881 \approx 6749.29
\]

### For $ t = 6 $ years:
\[
A = 5000 \times e^{0.06 \times 6}
\]
Calculating this:

\[
A \approx 5000 \times e^{0.36} \approx 5000 \times 1.43332941 \approx 7174.77
\]

Now, the corrected amounts to the nearest cent are:
- $ t = 1 $ year: **$5309.18**
- $ t = 2 $ years: **$5637.48**
- $ t = 3 $ years: **$5986.09**
- $ t = 4 $ years: **$6356.25**
- $ t = 5 $ years: **$6749.29**
- $ t = 6 $ years: **$7174.77**

Feel free to ask for more details or if you have any questions! 

Here are 5 related questions:
1. How does the formula for continuously compounded interest differ from other interest formulas?
2. What happens to the investment amount if the interest rate increases?
3. Can the same formula be used for monthly or quarterly compounded interest? Why or why not?
4. How does continuous compounding compare to simple interest over time?
5. What would the investment be after 10 years with continuous compounding?

**Tip**: Continuously compounded interest always grows faster than any other compounding frequency (like annual or monthly).

An investment of $5,000 is deposited into an account for which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the indicated times. (Round your answers to the nearest cent.) r = 6% Time (years) Amount 1 $ 5309.22 Incorrect: Your answer is incorrect. 2 $ 5637.50 Incorrect: Your answer is incorrect. See the rounding prompt for how many decimal places are needed. 3 $ 5970.25

Calculate the continuously compounded interest for a $5,000 investment at an interest rate of 6%. This problem demonstrates how to use the formula A = P * e^(rt) to find the investment amount after different time periods. Suitable for students in Grades 10-12, covering exponential growth and continuously compounded interest.

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