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 one year 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.)
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1looks like annuity setup

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

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

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0dw:1372927372461:dw

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0errr...my calculator shows that it's 165824.3866

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Why the long stuff?

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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0@primeralph this's a question about sequence

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0Well, both ways work. Mine is just shorter.

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0@primeralph but your equation does match the answer, neither @ganeshie8 ..

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

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@kryton1212 What's the answer?

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0172000 after corr. to 3 sig fig

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i mean the correct answer..

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

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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i don't know how to do... @hartnn show yourself_

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

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

kryton1212
 one year 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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0maybe @primeralph goes on

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0Uh well, tag away if you're satisfied.

ganeshie8
 one year 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.

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0at the beginning of 2013..................

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0@primeralph do you have any other ideas?

primeralph
 one year 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.

kryton1212
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2i am getting 172457.362022

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

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

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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0that means we use the wrong method @ganeshie8 .....

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0@hartnn million thanks. but how to present it in a shorter form...?

hartnn
 one year 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...

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0wow, i think this's more complicated..

hartnn
 one year 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...

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0uhh, anyway, thank you @hartnn
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