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Mathematics 15 Online
OpenStudy (anonymous):

Determine how much money will be in a savings account if the initial deposit was $750 and the interest rate is 2.465% compounded continuously for 20 years. (Round your answer to the nearest cent.)

OpenStudy (anonymous):

Here is the interest compounded continuously formula: p=p of e^rt

OpenStudy (anonymous):

seems straight forward but idk how to really do this

hero (hero):

Hold on a sec

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

u can use that program u used last time to help

hero (hero):

\[A = Pe^{rt}\] A = Balance after t years r = Annual Interest Rate

hero (hero):

P = Amount of Deposit

OpenStudy (anonymous):

yes

hero (hero):

\[A = (750)e^{(.02465)(20)}\]

hero (hero):

A = 1227.92

hero (hero):

Any questions?

OpenStudy (anonymous):

wheres the e on my calc

hero (hero):

What calculator are you using?

OpenStudy (anonymous):

TI 84

hero (hero):

I have TI-Nspire. You should get one of those. $50

hero (hero):

You have the e function on TI-84

hero (hero):

It's on there. You just have to find it

OpenStudy (anonymous):

ok i found it

OpenStudy (anonymous):

i have another similar question that says 15 years and 8 months... how would i punch that in? 15.8?

hero (hero):

8/12 = 2/3 = .66667

OpenStudy (anonymous):

so 15.66667?

OpenStudy (anonymous):

or wait no

OpenStudy (anonymous):

just .66667

hero (hero):

15.66667

OpenStudy (anonymous):

ok

hero (hero):

actually, it would be best if you use the fraction 47/3

hero (hero):

to be more exact

OpenStudy (anonymous):

ill try the decimal first and see

OpenStudy (anonymous):

it worked with the decimal just fine

OpenStudy (anonymous):

At an interest rate of 4% compounded annually, how long will it take to double the following investments? (Round your answer to the nearest year.) (a) $100 _____ years

OpenStudy (anonymous):

need help with this too

hero (hero):

Mario, you'll have to figure out some of these on your own. I'm going to be helping someone with their math

OpenStudy (anonymous):

...ok :(

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