Formula 1 point You deposit $$ \$ 112 $$ in an account that pays interest with annual compounding. If the account contains $$ \$ 145 $$ exactly 9 years from today, what is the annual interest rate? Enter your answer as a decimal, and show 4 decimal places. For example, if your answer is $$ 6.5 \% $$, you should enter .0650 . Type your answer.

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To find the annual interest rate, we can use the formula for compound interest:

$$ A = P(1 + r)^n $$

where:
- $$ A $$ is the amount of money accumulated after n years, including interest.
- $$ P $$ is the principal amount (initial deposit).
- $$ r $$ is the annual interest rate (in decimal form).
- $$ n $$ is the number of years the money is invested.

Given:
- $$ P = \$112 $$ (initial deposit)
- $$ A = \$145 $$ (amount after 9 years)
- $$ n = 9 $$ years

We need to solve for $$ r $$ (annual interest rate).

Substitute the given values into the formula:

$$ 145 = 112(1 + r)^9 $$

Now, we can solve for $$ r $$ using the formula for compound interest.
Solve the equation by following steps:
- step0: Solve for $$r$$:
  $$145=112\left(1+r\right)^{9}$$
- step1: Swap the sides:
  $$112\left(1+r\right)^{9}=145$$
- step2: Divide both sides:
  $$\frac{112\left(1+r\right)^{9}}{112}=\frac{145}{112}$$
- step3: Divide the numbers:
  $$\left(1+r\right)^{9}=\frac{145}{112}$$
- step4: Simplify the equation:
  $$\sqrt[9]{\left(1+r\right)^{9}}=\sqrt[9]{\frac{145}{112}}$$
- step5: Calculate:
  $$1+r=\sqrt[9]{\frac{145}{112}}$$
- step6: Simplify the root:
  $$1+r=\frac{\sqrt[9]{145\times 112^{8}}}{112}$$
- step7: Move the constant to the right side:
  $$r=\frac{\sqrt[9]{145\times 112^{8}}}{112}-1$$
- step8: Subtract the numbers:
  $$r=\frac{\sqrt[9]{145\times 112^{8}}-112}{112}$$
The annual interest rate $$ r $$ is approximately 0.0291 or 2.9108%.
The annual interest rate is 0.0291.

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