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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]

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  1. alexeis_nicole
    • 3 years ago
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    A 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?

  2. DominantShadow1
    • 3 years ago
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    darn im not this high in math...

  3. alexeis_nicole
    • 3 years ago
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    hm.. 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

  4. alexeis_nicole
    • 3 years ago
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    i used this because shouldn't i find the Present Value?

  5. imranmeah91
    • 3 years ago
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    \[12000\left(\frac{1-(1.089)^{-20}}{0.089}\right)\] 12000 because 1000 every month which means 12000 every year

  6. imranmeah91
    • 3 years ago
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    110328.

  7. alexeis_nicole
    • 3 years ago
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    hmm that's still not $111 943.89 though. :S sorry D:

  8. imranmeah91
    • 3 years ago
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    ok , getting there

  9. alexeis_nicole
    • 3 years ago
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    yep. for sure.

  10. imranmeah91
    • 3 years ago
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    what grade is that, btw?

  11. alexeis_nicole
    • 3 years ago
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    grade 11 financial applications (X

  12. imranmeah91
    • 3 years ago
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    are you sure your answer is absolutely correct, because I tried everything

  13. alexeis_nicole
    • 3 years ago
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    that'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

  14. alexeis_nicole
    • 3 years ago
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    really? :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.

  15. imranmeah91
    • 3 years ago
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    1 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.

  16. alexeis_nicole
    • 3 years ago
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    hm, 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

  17. alexeis_nicole
    • 3 years ago
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    ^20 **

  18. alexeis_nicole
    • 3 years ago
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    /(-R)*

  19. imranmeah91
    • 3 years ago
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    \[\left[12000\left(\frac{1-(r)^{-n}}{r}\right)\right.\]

  20. alexeis_nicole
    • 3 years ago
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    alright. 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 !

  21. imranmeah91
    • 3 years ago
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    np

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