An account earning interest at a rate of 4.6% per year, compounded continuously, will be worth $2174.30 in 18 years.  How much is invested in the account today, if no additional deposits or withdrawals are made?
Group of answer choices

$2.39

$950

$0.55

None of These

$4976.40

To solve this problem, we use the formula for continuously compounded interest:

\[
A = P e^{rt}
\]

Where:
- $ A $ is the amount in the account after $ t $ years (which is $2174.30).
- $ P $ is the principal, or the amount invested initially (which we need to find).
- $ r $ is the annual interest rate (4.6% or 0.046).
- $ t $ is the time the money is invested for (18 years).
- $ e $ is Euler's number, approximately equal to 2.71828.

Rearranging the formula to solve for $ P $:

\[
P = \frac{A}{e^{rt}}
\]

Now substitute the known values:

\[
P = \frac{2174.30}{e^{0.046 \times 18}}
\]

\[
P = \frac{2174.30}{e^{0.828}}
\]

Using $ e^{0.828} \approx 2.2892 $:

\[
P = \frac{2174.30}{2.2892} \approx 950
\]

Thus, the initial investment was approximately **$950**.

The correct answer is:

**$950**

---

Would you like more details on this calculation? Here are some related questions to explore:

1. What would the amount be in 25 years instead of 18 years?
2. How does the continuous compounding formula differ from annual compounding?
3. What is the investment worth if the interest rate changes to 5%?
4. How is Euler’s number derived and used in compounding?
5. What if the interest was compounded quarterly instead of continuously?

**Tip:** Continuously compounded interest grows faster than interest compounded at regular intervals (annually, monthly, etc.) due to the constant application of growth.

An account earning interest at a rate of 4.6% per year, compounded continuously, will be worth $2174.30 in 18 years. How much is invested in the account today, if no additional deposits or withdrawals are made?

Find out how much was initially invested in an account earning 4.6% interest compounded continuously, growing to $2174.30 over 18 years. This problem involves the continuous compound interest formula and exponential growth concepts. Suitable for high school students in Grades 10-12.

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