The formula for continuously compounded interest is:

$A = P \cdot e^{rt}$

Where:

• $A$ is the ending balance (future value),
• $P$ is the principal amount (initial deposit),
• $r$ is the interest rate (expressed as a decimal),
• $t$ is the time (in years),
• $e$ is the mathematical constant approximately equal to 2.71828.

In this case:

• $P = 10,000$,
• $r = 1.85\% = 0.0185$,
• $t = 3 \text{ years}$.

Now let's calculate the ending balance:

$A = 10,000 \cdot e^{0.0185 \cdot 3}$

I’ll compute this.The ending balance after three years with a continuously compounded interest of 1.85% would be approximately $10,570.69.

Would you like any further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the ending balance change if the interest rate were higher or lower?
2. What is the effect of compounding interest continuously versus annually or quarterly?
3. How does the formula for continuously compounded interest differ from simple interest?
4. What would the balance be if she left the money for 5 years instead of 3?
5. How can this formula be rearranged to find the interest rate given a future balance?

Tip: Continuously compounded interest grows faster than interest compounded annually because it's calculated and added to the balance at every moment.