If $$ \$ 450 $$ are deposited into an account with a $$ 8.5 \% $$ interest rate, compounded annually, what is the balance after 9 years? $$ F=\$[?] $$ $$ F=P\left(1+\frac{r}{n}\right)^{n t} $$

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To find the balance after 9 years, we can use the formula for compound interest:

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

Given:
- Principal amount ($$ P $$) = $450
- Interest rate ($$ r $$) = 8.5% = 0.085
- Compounding frequency ($$ n $$) = 1 (compounded annually)
- Time ($$ t $$) = 9 years

Substitute the given values into the formula:

$$ F = 450\left(1 + \frac{0.085}{1}\right)^{1 \times 9} $$

Now, we can calculate the balance after 9 years.
Calculate the value by following steps:
- step0: Calculate:
  $$450\left(1+0.085\right)^{9}$$
- step1: Add the numbers:
  $$450\times 1.085^{9}$$
- step2: Convert the expressions:
  $$450\left(\frac{217}{200}\right)^{9}$$
- step3: Simplify:
  $$450\times \frac{217^{9}}{200^{9}}$$
- step4: Rewrite the expression:
  $$50\times 9\times \frac{217^{9}}{200^{9}}$$
- step5: Rewrite the expression:
  $$50\times 9\times \frac{217^{9}}{50^{9}\times 4^{9}}$$
- step6: Reduce the numbers:
  $$9\times \frac{217^{9}}{50^{8}\times 4^{9}}$$
- step7: Multiply the fractions:
  $$\frac{9\times 217^{9}}{50^{8}\times 4^{9}}$$
The balance after 9 years is approximately $937.74.
The balance after 9 years is approximately $937.74.

If are deposited into an account
with a interest rate, compounded
annually, what is the balance after 9 years?

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If are deposited into an account with a interest rate, compounded annually, what is the balance after 9 years?