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The problem asks how much will be in an account after 10 years if $18,000 is deposited at an interest rate of 0.13% compounded annually. Let's calculate it using 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.
- $P = 18,000$ (the initial deposit).
- $r = 0.0013$ (since 0.13% = 0.0013 as a decimal).
- $n = 1$ (since the interest is compounded annually).
- $t = 10$ years.

Substituting these values into the formula:

\[
A = 18,000 \left(1 + \frac{0.0013}{1}\right)^{1 \times 10}
\]
\[
A = 18,000 \left(1 + 0.0013\right)^{10}
\]
\[
A = 18,000 (1.0013)^{10}
\]

Now, I will compute the value.After 10 years, the amount in the account will be approximately $18,235.37.

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

1. What would the amount be if the interest was compounded monthly instead of annually?
2. How much interest is earned after 10 years?
3. How would the final amount change if the interest rate was 1.3% instead of 0.13%?
4. What happens if the deposit time is increased to 20 years with the same interest rate?
5. How much should be deposited to reach $25,000 in 10 years with the same interest rate?

**Tip:** Small changes in interest rates can have a significant effect over long periods due to compounding.

$18,000 is deposited into an account that earns 0.13% interest compounded annually. How much is in the account after 10 years?

Learn how to calculate compound interest on a $18,000 deposit at a 0.13% annual interest rate over 10 years. Explore the compound interest formula and its application in this real-world problem.

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