To calculate the ending balance when $3,000 is invested at an interest rate of 1.8%, compounded hourly for two years, we can use the compound interest formula:

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

where:

• $P$ is the principal (initial amount) = $3,000,
• $r$ is the annual interest rate (in decimal) = 0.018,
• $n$ is the number of times the interest is compounded per year = 8,760 (since there are 24 hours in a day and 365 days in a year),
• $t$ is the time in years = 2.

Plugging in the values:

$A = 3000 \left(1 + \frac{0.018}{8760}\right)^{8760 \times 2}$

Now, calculating the expression step-by-step:

1. Calculate the hourly rate: $\frac{0.018}{8760}$.
2. Compute $8760 \times 2$.
3. Raise $\left(1 + \frac{0.018}{8760}\right)$ to the power of the result from step 2.
4. Multiply by 3,000 to get the final balance.

Let's proceed with these calculations:The ending balance after investing $3,000 at a 1.8% interest rate, compounded hourly for two years, is approximately $3,109.97.

Would you like a detailed breakdown of the calculations?

Here are some related questions for further exploration:

1. How does changing the compounding frequency to daily or monthly affect the ending balance?
2. What would be the final balance if the investment period were extended to five years?
3. How would the balance differ if the interest rate were increased to 3%?
4. How is the formula affected by different compounding periods, such as quarterly or annually?
5. What would the balance be if compounded continuously for the same period?

Tip: Compounding more frequently (such as hourly or daily) results in a slightly higher balance compared to annual compounding because interest is added more often.