A person invests 10000 dollars in a bank. The bank pays 6.75% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 37900 dollars?

it will take 20.6 years

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

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

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

A = $37900

P = $10000

r = 6.75% = 6.75/100 = 0.0675

n = 12 because it was compounded 12 times in a year.

Therefore,

37900 = 10000(1 + 0.065/12)^12 × t

37900/10000 = (1 + 0.00542)^12t

3.79 = (1.00542)^12t

Taking log of both sides, it becomes

Log 3.79 = 12t × log 1.00542

0.579 = 12t × 0.00234752004

0.579 = 0.02817t

t = 0.579/0.02817

t = 20.6 years

good for u

step-by-step explanation:

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer