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anonymous

  • one year ago

http://prntscr.com/8gesl4 Help would be great, thanks.

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  1. dm91
    • one year ago
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    i would say D but that is my opinion i would ask someone else

  2. dm91
    • one year ago
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    for there opinion

  3. anonymous
    • one year ago
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    Mmm, multiplying across seemed to have happened when the "u" was backwards, so would i have to do the opposite ? Or mmm idk i get lost with these "u's"

  4. dm91
    • one year ago
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    best one to ask would be steve816

  5. anonymous
    • one year ago
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    oh ok thanks @dm91 for your input though, and um @steve816 mind lending a quick hand with this problem, thanks.

  6. dm91
    • one year ago
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    anytime

  7. anonymous
    • one year ago
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    Ah algebra 2.. such a pain in my butt. Okay, let me see if I can type all of this out real quick because I just finished this class last year.

  8. anonymous
    • one year ago
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    yup such a pain ! and, oh ok thanks !

  9. anonymous
    • one year ago
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    So basically what we have here are intersections and unions. The intersection (upside down U) a set that contains elements that are common to both sets. Then a union (U) is a set that contains all elements of those 2 sets. If two events (A & B) are independent, then P(A [upside U] B) = P(A) times P(B). P (A U B) = P(A) + P(B) - P(A [upside down U] B). Since they give you P(A), P(B), and what P(A U B) is. It's all in the matter of plugging it in the equation.

  10. anonymous
    • one year ago
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    sooo.... it will be 2/3+4/5-8/15=P(AdownwarduB)

  11. anonymous
    • one year ago
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    i ended up with 14/15, lets give it ago ! see how it goes.

  12. anonymous
    • one year ago
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    @rheaanderson341 got it correct ! thanks for the help !

  13. anonymous
    • one year ago
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    Grr slow Google Chrome. Always welcome though! :)

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