David invested $220 in an account paying an interest rate of 1.7% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 10 years?

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

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

Where:
- $ A $ is the amount of money in the account after time $ t $.
- $ P $ is the principal amount (the initial amount of money).
- $ r $ is the annual interest rate (in decimal form).
- $ t $ is the time the money is invested or borrowed for, in years.
- $ e $ is the base of the natural logarithm, approximately equal to 2.71828.

Given:
- $ P = 220 $ dollars
- $ r = 1.7\% = 0.017 $
- $ t = 10 $ years

Substituting these values into the formula:

\[
A = 220 \cdot e^{0.017 \cdot 10}
\]

\[
A = 220 \cdot e^{0.17}
\]

Now we calculate $ e^{0.17} $:

\[
e^{0.17} \approx 1.1855
\]

Then:

\[
A \approx 220 \cdot 1.1855 \approx 260.81
\]

To the nearest ten dollars, the amount in the account after 10 years is approximately **$260**.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How does the final amount change if the interest rate is increased to 2%?
2. What would be the amount after 20 years with the same interest rate?
3. How much interest is earned after 10 years?
4. What is the effect of compounding semi-annually instead of continuously on the final amount?
5. How long would it take for the investment to double at the same interest rate?

**Tip:** Continuous compounding results in a slightly higher amount compared to other compounding methods like annual or monthly compounding because it assumes the interest is constantly being added to the principal.

Explore how to calculate compound interest with continuous compounding using David's investment of $220 at an interest rate of 1.7% over 10 years. Learn the formula for continuous compounding and see a detailed step-by-step solution. Suitable for advanced high school students.

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