Raymond B. answered 04/25/23
Math, microeconomics or criminal justice
8058=6000(1+.07/4)^4t
8058/6000=4029/3000= 1342/1000= 1.342= 1.0175^4t
take logs to base 1.0175
4t= ln1.342/ln1.0175
t = about4.239 years
= about 4.24
Nicole D.
asked 04/25/23Boris needs $8058 for a future project. He can invest $6000 now at an annual rate of 7%, compounded quarterly. Assuming that no withdrawals are made, how long will it take for him to have enough money for his project?
Do not round any intermediate computations, and round your answer to the nearest hundredth.
Raymond B. answered 04/25/23
Math, microeconomics or criminal justice
8058=6000(1+.07/4)^4t
8058/6000=4029/3000= 1342/1000= 1.342= 1.0175^4t
take logs to base 1.0175
4t= ln1.342/ln1.0175
t = about4.239 years
= about 4.24
A = P(1+r/n)nt
8058 = 6000(1+0.07/4)4t
8058/6000 = (1+0.0175)4t
ln(8058/6000) = 4t*ln(1.0175)
ln(8058/6000)/(4ln(1.0175)) = t
t ≈ 4.25
Thus, it will take Boris 4.25 years to have enough money for his project.
Hope this helped!
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