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alexeis_nicole
 3 years ago
could someone please help me find my R value for this equation? i'm not quite sure as to what it should be to solve for the equation PV= R [1(1+i)^n/i]
alexeis_nicole
 3 years ago
could someone please help me find my R value for this equation? i'm not quite sure as to what it should be to solve for the equation PV= R [1(1+i)^n/i]

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alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1A lottery to raise a funds for a hospital is advertising a $240 000 prize. The winner will receive $1000 ever month for 20 years., starting a year from now. a) if the interest is 8.9% per annum, compounded annually, how much must be invested not to have the money to pay this prize?

DominantShadow1
 3 years ago
Best ResponseYou've already chosen the best response.0darn im not this high in math...

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1hm.. the way i did it was like this \[240000\left[ 1(1.089)^{20} / 0.089 \right]\] but i got the wrong answer. the answer should be $111 943.89 but... i haven't really gotten that yet.. :S

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1i used this because shouldn't i find the Present Value?

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.1\[12000\left(\frac{1(1.089)^{20}}{0.089}\right)\] 12000 because 1000 every month which means 12000 every year

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1hmm that's still not $111 943.89 though. :S sorry D:

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.1what grade is that, btw?

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1grade 11 financial applications (X

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.1are you sure your answer is absolutely correct, because I tried everything

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1that's what the answer at the back of the book says. and yeah.. im trying different possibilities as well.. but nothing's been working for me either

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1really? :S how did you get that? and yea i know for a fact that i can't trust some of the answers in the back of the book.

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.11 st Year = PV 2nd Year= PV(1+R)12000 3rd Year= (PV (1 + R)^2  12000 (1 + R))  12000 21th Year =(PV(1+R)^20) 12000((1+R)^0+(1+R)^1+....(1+R)^19) rewrite using geometric sum \[\text{PV}(1+R)^{20}12000\frac{\left(1(1+R)^{20}\right)}{(R)}\text{==}0\] we know R, so solve for PV and I get PV = 110328.

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1hm, that seems to be the closest the answer could get to 111 943. 89 so i guess i'll stick with it for now until i get back to school. i'll ask my teacher if there was a typo in the answer or something.. so the equation is PV(1+R)20−12000(1−(1+R)20)(−R)=0 ? right? R=280000

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.1\[\left[12000\left(\frac{1(r)^{n}}{r}\right)\right.\]

alexeis_nicole
 3 years ago
Best ResponseYou've already chosen the best response.1alright. i'll just keep it like this for now. and then when i get back to school i'll ask my teacher. thanks for the help imran !
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