a person invests 3000 dollars in a bank. the bank pays 5.75% interest compounded annually. to the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 7200 dollars?

15.7 years

Step-by-step explanation:

Use the formula for compound interest. A = P(1 + i)ⁿ

A is the total amount of money. A = 7200

P is the principal, starting money. P = 3000

i is the interest per compounding period in decimal form. Since interest is compounded annually, i = 0.0575

n is the number of compounding periods. n = ?

Substitute the information into the formula and isolate n.

A = P(1 + i)ⁿ

7200 = 3000(1 + 0.0575)ⁿ    Solve inside the brackets

7200 = 3000(1.0575)ⁿ

7200/3000 = 1.0575ⁿ       Divide both sides by 3000

2.4 = 1.0575ⁿ

n = (㏒ ans) / (㏒ base)

n = (㏒ (2.4)) / (㏒ (1.0575))

n = 15.659     Exact answer

n ≈ 15.7       Rounded to the nearest tenth of a year

Therefore the person must leave the money in the bank for 15.7 years until it reaches 7200 dollars.

answer is:

1.75 pounds

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p