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Homework Help: Question 19 - quadratic probility problem

  1. Apr 4, 2007 #1
    img021.jpg

    a) (i) (x-14)(2x-7)
    (ii) x = 14 or x = 3.5

    b)
    i) [tex]\frac{7}{n+7}[/tex]
    ii) Take n to be 8

    [tex]\frac{7}{8+7}[/tex]
    [tex]\frac{7}{15}[/tex] that DOESN'T round down to [tex]\frac{2}{5}[/tex]

    Is that all correct so far?
    If so I will post the next (really hard) question)...

    Thanks
     
  2. jcsd
  3. Apr 4, 2007 #2

    HallsofIvy

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    Why take n to be 8? Nothing is said about any value for n.
    If 7/(n+7)= 2/5, SOLVE for n. What happens?
     
  4. Apr 4, 2007 #3
    [tex]\frac{7}{n+7}[/tex] = [tex]\frac{2}{5}[/tex]

    cross multiply!

    [tex]\frac{35}{2n+14}[/tex]

    now where?
     
  5. Apr 4, 2007 #4
    Um...you're leaving out the equality part of the equation. Solve for n.
     
  6. Apr 4, 2007 #5
    Alternatively, you can note that you have 35/(2n + 14) = 1. 35 is odd. 2n + 14 is even. Strange, isn't it?
     
  7. Apr 4, 2007 #6
    uh? How does that equal 1?
     
  8. Apr 4, 2007 #7
    Look, you're essentially supposed to say, "Suppose Bill is right. Suppose 7/(n + 7) = 2/5. Then such and such would follow." Why would the conclusion be a problem?
     
  9. Apr 4, 2007 #8
    [tex]\frac{35}{2n+14}[/tex] = 1

    now I need to get N on it's own (don't know how- plz show). But im guessing that n is greater than 3/5 so it CANT be right?
     
  10. Apr 4, 2007 #9

    danago

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    Multiply both sides of the equation by 2n + 14, so you end up with:

    35 = 2n + 14
     
  11. Apr 4, 2007 #10
    silly me....

    2n = 35-14
    n = 10.5
    EDIT: Which as a fraction is 10/1/2 which DOESN'T equal 2/5...am i right yet. I doubt thats right...
     
  12. Apr 4, 2007 #11
    You're getting there. What's the problem with n being 10.5. Look at your original assumptions. What are you tacitly assuming about the original n balls?
     
  13. Apr 5, 2007 #12
    you cant have 1/2 a ball... :bugeye:

    am I right or am I right
     
    Last edited: Apr 5, 2007
  14. Apr 6, 2007 #13
    Okay here is the rest of the question

    img022.jpg
     
  15. Apr 6, 2007 #14

    Hootenanny

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    Okay, what are your ideas on part (c)?
     
  16. Apr 6, 2007 #15
    (probability of taking white x probability of taking yellow) + (probability of taking yellow x probality of taking white)

    Is that somthing to go from?

    Thanks
     
  17. Apr 6, 2007 #16

    Hootenanny

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    Yes, I'd go with that.
     
  18. Apr 6, 2007 #17
    I got it down now to [tex]\frac{14n}{2n^{2}+28n+98}[/tex] = [tex]\frac{4}{9}[/tex]

    is that right so far?

    EDIT: is the question is says -28n but ive got +28n


    EDIT 2: O no it must be this so far

    [tex]\frac{14n}{n^{2}+14n+49}[/tex] = [tex]\frac{4}{9}[/tex]
     
  19. Apr 6, 2007 #18
    got it!

    [tex]\frac{4n^{2} + 56n + 196}{2}[/tex] = [tex]\frac{14n * 9}{2}[/tex]
    [tex]\2n^{2} + 28n + 98[/tex] = [tex]68n[/tex]


    [tex]\2n^{2} - 35n + 98 = 0[/tex]
     
  20. Apr 6, 2007 #19
    and the answer to d must be 1/9
    x must be 14 as you cant have 3.5 balls
     
  21. Apr 6, 2007 #20

    Hootenanny

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    All sounds good to me, well done :approve:
     
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