Find the ending balance in an account that opens with $6,000, earns 4.5% interest compounded quarterly, and is held for 10 years. (Round your answer to the nearest cent.)

Pt = Po(1+r)^n.

r=(4.5%/4) / 100% = 0.01125=Quarterly %
rate expressed as a decimal.

n = 4comp./yr * 10yrs = 40 Compounding
periods.

Pt = $6000(1.01125)^40 = $9386.26.

To find the ending balance in the account, we'll use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A = the ending balance (or future value) of the account
P = the principal amount (initial deposit), in this case $6,000
r = the annual interest rate (4.5%)
n = the number of times interest is compounded per year (quarterly, so 4 times)
t = the number of years the money is invested (10 years)

Plugging the values into the formula, we get:

A = 6000(1 + 0.045/4)^(4*10)

Now, let's calculate it step by step:

Step 1: Divide the annual interest rate by the number of compounding periods per year:
r/n = 0.045/4 = 0.01125

Step 2: Add 1 to the result:
1 + r/n = 1 + 0.01125 = 1.01125

Step 3: Multiply the number of compounding periods per year by the number of years:
nt = 4 * 10 = 40

Step 4: Raise the result from Step 2 to the power of the result from Step 3:
(1 + r/n)^(nt) = (1.01125)^40

Using a calculator, we find that (1.01125)^40 is approximately 1.214649...

Step 5: Multiply the principal amount by the result from Step 4 to obtain the ending balance:
A = 6000 * 1.214649 ≈ $7,287.89

Therefore, the ending balance in the account after 10 years would be approximately $7,287.89.