A U.S. citizen invested a $1200 government stimulus check they received on March 1, 2020 into stocks that grew with an annual percentage rate of 12.3%, compounded continuously. How much are these stocks worth 6 months later on September 1, 2020? Remember to round your answer to two decimal place accuracy. (Hint: 6 months = 0.5 years.)

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

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

Where:
- $ A $ is the amount of money accumulated after interest,
- $ P $ is the principal (initial investment),
- $ r $ is the annual interest rate (expressed as a decimal),
- $ t $ is the time the money is invested or borrowed for, in years,
- $ e $ is the base of the natural logarithm.

### Given:
- $ P = 1200 $ (the initial investment),
- $ r = 12.3\% = 0.123 $ (the annual percentage rate as a decimal),
- $ t = 0.5 $ (6 months = 0.5 years).

Now, let's calculate the final amount $ A $.

\[
A = 1200 \cdot e^{0.123 \cdot 0.5}
\]

I'll compute this now.The value of the stocks after 6 months, on September 1, 2020, would be approximately **$1276.12**, when rounded to two decimal places.

Would you like further details or have any questions? Here are 5 related questions for practice:

1. What would the value be if the investment period was 1 year?
2. How would the outcome change if the interest rate was compounded monthly instead of continuously?
3. What is the formula for compound interest when compounding is done annually instead of continuously?
4. If the interest rate were increased to 15%, how would that affect the final value after 6 months?
5. How long would it take for the investment to double at the same interest rate?

**Tip:** Continuous compounding uses the constant $ e $, making it slightly more profitable than discrete compounding methods like monthly or yearly.

Learn how to calculate the value of a $1200 investment after 6 months, with a 12.3% annual interest rate compounded continuously. This problem covers exponential growth, the continuous compound interest formula, and natural logarithms. Suitable for high school students in Grades 10-12.

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