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anonymous
 2 years ago
Mr. Chan will deposit $25000 in a bank account at the beginning of each year from 2007 to 2012. If the interest is compounded yearly and the interest rate is 4% p.a., find the total amount that he can get at the beginning of 2013. (corr. to 3 sig. fig.)
anonymous
 2 years ago
Mr. Chan will deposit $25000 in a bank account at the beginning of each year from 2007 to 2012. If the interest is compounded yearly and the interest rate is 4% p.a., find the total amount that he can get at the beginning of 2013. (corr. to 3 sig. fig.)

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ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1looks like annuity setup

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1P = 25,000 r = .04 n = 6

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1\(A = P[ (1+r)^n + (1+r)^{n1} + .... (1+r)]\) = \(P[\frac{1(1+r)^n }{ 1(1+r)}]\)

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1ha take ur time not an easy problem :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1372927372461:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0errr...my calculator shows that it's 165824.3866

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Why the long stuff?

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1every year he is adding 25 grand

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@primeralph this's a question about sequence

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Well, both ways work. Mine is just shorter.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@primeralph but your equation does match the answer, neither @ganeshie8 ..

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1prime i used ur formula and took a sum to arrive at the final formula

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@kryton1212 What's the answer?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0172000 after corr. to 3 sig fig

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i mean the correct answer..

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0It is not possible to get that much money after just 6 years.

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=25000%281%281%2B.04%29%5E6%29%2F%281%281%2B.04%29%29

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1im getting 165,824 @kryton1212 prime read the q again

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i don't know how to do... @hartnn show yourself_

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1which formula wont work ?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh..........beginning of each year. Hold on then.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i also think that @ganeshie8 's formula is correct...maybe i should ask my teacher about this question. Let's move on to the next question that you both will be tagged by me a moment later

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0maybe @primeralph goes on

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Uh well, tag away if you're satisfied.

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1move on to new problem, im pretty sure what we worked is correct. if you do manually 6 times also, you would get the same answer.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0at the beginning of 2013..................

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@primeralph do you have any other ideas?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0To check, you can do each step. If you get the same thing, let it go.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i am lazy XD...i will ask my teacher about this question tonight. thank you both

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2i am getting 172457.362022

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.13 sig figs, thats the right answer also... how ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2i calculated for each year...6 calculations :\

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2200825k(1+0.04)=26k 2009(26k+25k)(1+0.04)=53040 2010(53040+25k)(1+0.04)=81161.1 2011(81161.1+25k)(1+0.04)=110408.064 2012(110408.064+25k)(1+0.04)=140824.38 2013(140824.38+25k)(1+0.04)=172457....

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1oh every year rounding off to 3 sigfigs is it

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0that means we use the wrong method @ganeshie8 .....

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@hartnn million thanks. but how to present it in a shorter form...?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2thats what i am thinking... if we have it for just 2 years, \(P(1+r)+ [P(1+r)+P](1+r) = P [(1+r)+(1+r)(2+r)]\) so, maybe for 6 years, the formula will become \(P (1+r)[1+(2+r)[1+(3+r)[1+(4+r)[1+(5+r)[1+(6+r)]]]]]\) maybe there's a simplification involved, maybe...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wow, i think this's more complicated..

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2hmm.....maybe they wanted us to calculate each year's separately...because i don't see any explicit formula for this...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0uhh, anyway, thank you @hartnn
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