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kryton1212
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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
 one year ago
kryton1212 Group Title
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
 one year ago

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kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
@hartnn
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
looks like annuity setup
 one year ago

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

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
plug and chug..
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ahhh....wait
 one year ago

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

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

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

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

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

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

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

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

primeralph Group TitleBest ResponseYou've already chosen the best response.0
Uh hold on.
 one year ago

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

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

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
172000 after corr. to 3 sig fig
 one year ago

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

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

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

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

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

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

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

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i 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
 one year ago

primeralph Group TitleBest ResponseYou've already chosen the best response.0
Hold on.
 one year ago

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

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
move 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.
 one year ago

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

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

primeralph Group TitleBest ResponseYou've already chosen the best response.0
To check, you can do each step. If you get the same thing, let it go.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i am lazy XD...i will ask my teacher about this question tonight. thank you both
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i am getting 172457.362022
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ohh, how?
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
3 sig figs, thats the right answer also... how ?
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
200825k(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....
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
oh every year rounding off to 3 sigfigs is it
 one year ago

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
thats 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...
 one year ago

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
hmm....welcome ^_^
 one year ago
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