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

$A = P(1 + \frac{r}{n})^{nt}$

Where:

• $A$ is the amount in the account after interest.
• $P$ is the principal (initial deposit).
• $r$ is the annual interest rate (as a decimal).
• $n$ is the number of times the interest is compounded per year.
• $t$ is the time the money is invested or borrowed for, in years.

Given:

• $P = 5000$ (initial deposit)
• $r = 9.8\% = 0.098$ (annual interest rate as a decimal)
• $n = 12$ (monthly compounding)
• $t = 1$ year

Now, let's plug these values into the formula:

$A = 5000 \left( 1 + \frac{0.098}{12} \right)^{12 \times 1}$

First, calculate the inner expression:

$\frac{0.098}{12} = 0.0081667$

So the formula becomes:

$A = 5000 \left( 1 + 0.0081667 \right)^{12}$ $A = 5000 \left( 1.0081667 \right)^{12}$

Next, calculate $1.0081667^{12}$:

$1.0081667^{12} \approx 1.102576$

Now, multiply by 5000:

$A = 5000 \times 1.102576 = 5512.88$

Final Answer:

The total amount in the account after one year is approximately $5,512.88.

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Would you like further explanation or have any questions?

Here are 5 follow-up questions to expand on this:

1. How does the number of compounding periods affect the total amount in an account?
2. What would happen to the total amount if the interest rate were higher or lower?
3. How would the total amount change if the interest were compounded quarterly instead of monthly?
4. What is the difference between compound interest and simple interest?
5. How do you calculate the principal if you know the amount, interest rate, and time?

Tip: The more frequently the interest is compounded, the more interest you will earn, so monthly compounding (instead of annually) results in a higher final amount.