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mmend98

  • one year ago

Diana lost 35 points on a project for school. She can make changes and receive three-tenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.

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  1. triciaal
    • one year ago
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    |dw:1439420130504:dw|

  2. mmend98
    • one year ago
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    is there any way you could type that haha its very hard to interpret this. but thank you @triciaal

  3. Nnesha
    • one year ago
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    i'll type it :=) \[\huge\rm S_n =a+ar^{n-1}\] \[\large \rm (\frac{ 3 }{ 10 })^9 =a(1+r^{n-1})\] \[\large \rm 35 =\frac{ a(1+(\frac{ 3 }{ 10 })^9) }{ 1 -\frac{3}{10} }\] (formula: \[S_n = \frac{ a(1-r^n) }{ 1-r }\])

  4. mmend98
    • one year ago
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    So how would I plug this in?

  5. mmend98
    • one year ago
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    @triciaal

  6. mmend98
    • one year ago
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    Help ! haha please good person @ganeshie8

  7. ganeshie8
    • one year ago
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    triciaal seems to be having the right method - i think you just need to add up the first 10 terms

  8. ganeshie8
    • one year ago
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    remember the "partial sum formula" for geometric series ?

  9. mmend98
    • one year ago
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    yes i do

  10. ganeshie8
    • one year ago
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    \[\large S_n=a_1*\dfrac{1-r^n}{1-r}\] does it look something like above in ur notes ?

  11. mmend98
    • one year ago
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    yes. but there is the sigma thing in the front

  12. ganeshie8
    • one year ago
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    \[\large \sum\limits_{i=1}^na_i=a_1*\dfrac{1-r^n}{1-r}\] like that ?

  13. mmend98
    • one year ago
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    yes

  14. mmend98
    • one year ago
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    ok one sec

  15. ganeshie8
    • one year ago
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    No wait, there is a mistake

  16. mmend98
    • one year ago
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    ok

  17. ganeshie8
    • one year ago
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    scratch that, il type it again fixing the mistake

  18. mmend98
    • one year ago
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    ok thanks

  19. ganeshie8
    • one year ago
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    Okay, good. Lets see whats going on in the given word problem. She can make changes and receive three-tenths of the missing points back, so this becomes the common ratio : \(r=\dfrac{3}{10}\) To start with, the missing points were 35, so the first term must be : \(a_1 = 35*\dfrac{3}{10} = 10.5\) Plug them in the partial sum formula and evaluate

  20. ganeshie8
    • one year ago
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    \[\large \sum\limits_{i=1}^{10}a_i=10.5*\dfrac{1-\left(\frac{3}{10}\right)^{10}}{1-\frac{3}{10}}\] simplify..

  21. mmend98
    • one year ago
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    yes thank you

  22. ganeshie8
    • one year ago
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    let me know what you get and please medal triciaal, not me

  23. mmend98
    • one year ago
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    I got 149

  24. ganeshie8
    • one year ago
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    Wow! 149 ? She cannot receive 149 points back! double check ur work

  25. mmend98
    • one year ago
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    oh sorry put it in way wrong haha

  26. mmend98
    • one year ago
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    let me do it over

  27. ganeshie8
    • one year ago
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    take your time..

  28. mmend98
    • one year ago
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    wait yea i got 149 again lol

  29. ganeshie8
    • one year ago
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    lets check with wolfram

  30. mmend98
    • one year ago
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    ok

  31. ganeshie8
    • one year ago
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    http://www.wolframalpha.com/input/?i=10.5*%281-%283%2F10%29%5E10%29%2F%281-3%2F10%29

  32. mmend98
    • one year ago
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    ahh ok

  33. ganeshie8
    • one year ago
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    wolf says\(14.9999\), which approximates to \(15\) looks you have just missed one decimal point somewhere..

  34. mmend98
    • one year ago
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    but doesn't the sigma change it to 149

  35. ganeshie8
    • one year ago
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    why ?

  36. ganeshie8
    • one year ago
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    what do you know about sigma

  37. mmend98
    • one year ago
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    like when u account for sigma in the problem.

  38. ganeshie8
    • one year ago
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    what do you mean by that, could you elaborate

  39. mmend98
    • one year ago
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    yea you have to take in to account the numbers above the sigma

  40. mmend98
    • one year ago
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    idk maybe I'm crazy haha

  41. ganeshie8
    • one year ago
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    \(\sum\) is just a short cut for "sum of terms" don't let that confuse you here

  42. ganeshie8
    • one year ago
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    check this quick http://www.mathsisfun.com/algebra/sigma-notation.html

  43. mmend98
    • one year ago
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    ohh thanks so much

  44. ganeshie8
    • one year ago
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    so how many points is she gonna get back after 10 attempts ? whats the maximum possible points that she can get back if she is allowed to attempt infinitely many times ?

  45. mmend98
    • one year ago
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    15 points !

  46. ganeshie8
    • one year ago
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    careful here, \(14.999\) is not exactly same as \(15\) In fact, she can never get back \(15\) points even if she is allowed to attempt 100 or 1000 or any other big finite number of times

  47. mmend98
    • one year ago
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    ah. so it would just be 14.999

  48. mmend98
    • one year ago
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    thanks again

  49. ganeshie8
    • one year ago
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    She gets back 14.999 after 10 attempts The maximum points that she can get back is 15 if she is allowed to try "infinite" times (which we know is practically impossible here)

  50. ganeshie8
    • one year ago
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    Notice that with each attempt, the sum is by increasing a tiny bit and getting closer to 15, but it never quite reaches 15. We say the sum "converges" to 15.

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