\begin{tabular}{l} Suppose that $$ \$ 9000 $$ is placed in an account that pays $$ 6 \% $$ interest compounded each year \ Assume that no withdrawals are made from the account. \ Follow the instructions below. Do not do any rounding. \ (a) Find the amount in the account at the end of 1 year. \ $$ \$ 9540 $$ \ \hline (b) Find the amount in the account at the end of 2 years. \ $$ \$ 10112.4 $$\end{tabular}

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To find the amount in the account at the end of each year with compound interest, we can use the formula:

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

where:
- $$ A $$ is the amount in the account after $$ n $$ years,
- $$ P $$ is the principal amount (the initial amount of money),
- $$ r $$ is the annual interest rate (as a decimal),
- $$ n $$ is the number of years the money is invested or borrowed.

Given:
- $$ P = 9000 $$
- $$ r = 0.06 $$ (which is $$ 6\% $$)
- $$ n $$ will be $$ 1 $$ for part (a) and $$ 2 $$ for part (b).

### (a) Amount at the end of 1 year:

Using the formula for $$ n = 1 $$:

$$
A = 9000(1 + 0.06)^1
$$
$$
A = 9000(1.06)
$$
$$
A = 9000 \times 1.06 = 9540
$$

So, the amount in the account at the end of 1 year is $$ \$ 9540 $$.

### (b) Amount at the end of 2 years:

Using the formula for $$ n = 2 $$:

$$
A = 9000(1 + 0.06)^2
$$
$$
A = 9000(1.06)^2
$$
Calculating $$ (1.06)^2 $$:

$$
(1.06)^2 = 1.1236
$$

Now substituting back into the formula:

$$
A = 9000 \times 1.1236 = 10112.4
$$

So, the amount in the account at the end of 2 years is $$ \$ 10112.4 $$.

### Summary:
- (a) Amount at the end of 1 year: $$ \$ 9540 $$
- (b) Amount at the end of 2 years: $$ \$ 10112.4 $$
(a) Amount at the end of 1 year: $$ \$ 9540 $$ (b) Amount at the end of 2 years: $$ \$ 10112.4 $$

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