An engineer places $7,000 at the end of every year into a retirement account for 22 years. If the account into which the savings was placed earns 9% per year, how much was in the account at the end of the engineer's career? Express your answer in $ to the nearest $1,000.

$440,113.37

Explanation:

Since the engineer is placing $7,000 at the end of every year for the 22 years, therefore the amount which will be saved by him at the end of 22 years  shall be determined through the future value of annuity formula which is given as follows:

Amount after 22 years=R[((1+i)^n-1)/i]

In the given question

R=amount saved by engineer per year=$7,000

i=interest rate involved=9%

n=number of payment to be made=22

Amount after 22 years=7,000[((1+9%)^22-1)/9%]

                                    =$440,113.37

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