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Mathematics 15 Online
OpenStudy (anonymous):

What is A if r = 0.005, t = 300, and P = $2500.00? Use A = P(1 + r)t. A. $2,530.89 B. $10,780.45 C. $11,162.42 D. $12,172.35

OpenStudy (anonymous):

don't you have a calculator?

OpenStudy (mathmale):

Dear Best In The West, Why not substitute the given values for r, t and P into the given equation? Type out for us what you're doing, so that someone can give you appropriate feedback on your work.

OpenStudy (anonymous):

Well... first you put the number in the places in which they are supposed to be placed..

OpenStudy (anonymous):

Do u know what it would look like?

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

A= P(1+r)t A=2,500(1+.005) * 300

OpenStudy (anonymous):

Remember PEMDAS?

OpenStudy (anonymous):

lol, yes.

OpenStudy (anonymous):

okay so use it.

OpenStudy (anonymous):

what goes first?

OpenStudy (phi):

I think he means A= P(1+r)^t (and not A= P(1+r)t) this is an equation for finding how something grows exponentially ( like money in the bank) at a rate r

OpenStudy (mathmale):

Dear @bestinthewest : Mind double-checking to ensure that you've copied this problem down correctly: What is A if r = 0.005, t = 300, and P = $2500.00? Use A = P(1 + r)t. Is t really 300 years? None of us will live that long. :) Also, r=0.005 signifies 1/2 of one percent. That's a bit more likely. Lastly, did you mean A=P(1+r)t or \[A=P(1+r)^{t}\] I strongly suspect that you meant the latter.

OpenStudy (mathmale):

Supposing that P = 2500, r = 0.01, and t = 5 years, then after 5 years of compounding annually at the rate 1%, you would end up with \[A=2500(1+.01)^{5}\] Hope this example helps you to solve the problem given you.

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