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Lucy invested $19,000 in an account paying an interest rate of 3% compounded annually. Assuming no deposits or withdrawals are made, how much money, *to the nearest hundred dollars*, would be in the account after 8 years?

Answer

Attempt 1 out of 2

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

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $ A $ is the amount of money accumulated after $ t $ years, including interest.
- $ P $ is the principal amount (the initial money invested).
- $ r $ is the annual interest rate (decimal).
- $ n $ is the number of times that interest is compounded per year.
- $ t $ is the number of years the money is invested or borrowed for.

Given:
- $ P = 19,000 $ dollars
- $ r = 3\% = 0.03 $
- $ n = 1 $ (since the interest is compounded annually)
- $ t = 8 $ years

The formula simplifies to:

\[
A = 19,000 \left(1 + \frac{0.03}{1}\right)^{1 \times 8}
\]

\[
A = 19,000 \left(1 + 0.03\right)^{8}
\]

\[
A = 19,000 \times (1.03)^8
\]

Now, we calculate:

\[
A = 19,000 \times 1.26677 \approx 24,068.63
\]

Rounding this to the nearest hundred dollars:

\[
A \approx 24,100
\]

**Final Answer:** After 8 years, Lucy would have approximately **$24,100** in the account.

Would you like more details on any part of this solution? Here are some additional questions you could explore:

1. What if the interest were compounded quarterly instead of annually?
2. How would the outcome change if the interest rate were 4% instead of 3%?
3. What would be the balance after 5 years instead of 8 years?
4. How much interest would be earned in total after 8 years?
5. What is the formula for continuous compounding, and how would it apply here?

**Tip:** When rounding, remember to consider whether you're rounding up or down based on the decimal value.

Learn how to calculate compound interest after 8 years with an initial investment of $19,000 at an annual interest rate of 3%. Understand the formula A = P (1 + r/n)^(nt) and see step-by-step calculations to find the final amount. Suitable for Grades 10-12.

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