Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Please help me with this! What principal invested at 12% compounded continuously for 4 years will yield $1470? Round the answer to two decimal places. a. $2375.63 b. $1468.38 c. $562.85 d. $909.61

OpenStudy (tkhunny):

Have you considered evaluating: \(1470 = Deposit\cdot e^{0.12\cdot 4}\) I do hope it is obvious that the first two are no good. (a) is just silly. (b) is too close.

OpenStudy (wolf1728):

Principal = Total / (1 + rate)^years

OpenStudy (anonymous):

Okay, thank you guys!

OpenStudy (anonymous):

I got 909.61!

OpenStudy (anonymous):

Is that right?

OpenStudy (wolf1728):

cats - I'm looking for a formula for continuously compounded interest - be juat a few minutes.

OpenStudy (tkhunny):

If you didn't do anything wrong deliberately, there is a good chance you got it right. What do you think? Did you get it right?

OpenStudy (anonymous):

Great! I think I got it right, thanks for the help guys!

OpenStudy (tkhunny):

Perfect. A little confidence goes a long way. Good work.

OpenStudy (wolf1728):

I got the formula continuously compounded interst rate = e^(annual rate) -1 where e = 2.718281828 so 12% continuously = ((2.718281828)^.12)-1 =.1274968516

OpenStudy (anonymous):

Okay, thank you so much! =)

OpenStudy (wolf1728):

I've got the formula worked out too. Principal = Total / (1 + rate)^years Principal = 1,470 / (1.12)^4 Principal = 909.61 Here's a handy calculator to check your work: http://www.1728.org/compint.htm

OpenStudy (anonymous):

Great, thanks for all the help!

OpenStudy (wolf1728):

Thank you :-)

OpenStudy (tkhunny):

Ummm.... \(\dfrac{1470}{1.12^{4}} = 934.22 \ne 909.61\) is Annual Compounding Seemed kind of magic, up there. Had you used the rather silly .1274968516. Seriously, just use the exponential functions. There is absolutely no need to convert to annual interest.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!