If $$ \$ 650 $$ are deposited into an account with a $$ 4 \% $$ interest rate, compounded annually, what is the balance after 18 years? $$ F=\$[?] $$ $$ F=P\left(1+\frac{1}{n}\right)^{n t} $$

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To calculate the future balance $$ F $$ of an account with an initial deposit, compounded annually, you can use the compound interest formula:

$$
F = P \left(1 + r\right)^t
$$

Where:
- $$ P $$ is the principal amount (initial deposit).
- $$ r $$ is the annual interest rate (in decimal).
- $$ t $$ is the time the money is invested for, in years.

Given:
- $$ P = \$650 $$
- $$ r = 4\% = 0.04 $$
- $$ t = 18 $$ years

Plugging these values into the formula:

$$
F = 650 \times \left(1 + 0.04\right)^{18}
$$

$$
F = 650 \times (1.04)^{18}
$$

Now, let's compute $$ (1.04)^{18} $$:

$$
(1.04)^{18} \approx 2.0258
$$

Multiplying this by the principal:

$$
F \approx 650 \times 2.0258 \approx 1,316.77
$$

**Final Balance:**

$$
F \approx \$1,\!316.78
$$

So, after **18 years**, the account balance will be approximately **\$1,316.78**.
After 18 years, the balance will be approximately \$1,316.78.

If \( \$ 650 \) are deposited into an account with a \( 4 \% \) interest rate, compounded annually, what is the balance after 18 years? \[ F=\$[?] \] \( F=P\left(1+\frac{1}{n}\right)^{n t} \)

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