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anonymous

  • 5 years ago

2, 3, and 5 are substituted at random for a, b, and c in the equation ax+b=c. Probability that solution is negative? P( absolute value = 1?

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  1. anonymous
    • 5 years ago
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    ok, well you can brute force this

  2. anonymous
    • 5 years ago
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    if assignment is random, then the probability of something is always (number of success possibilities)/ (total number of possibilities)

  3. anonymous
    • 5 years ago
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    so there are a total of six possibilities and we can actually list them all:

  4. anonymous
    • 5 years ago
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    hmmm actually ok misread, there is an x variable haha

  5. anonymous
    • 5 years ago
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    ok so let's list the six possibilities anyway and see where that gets us

  6. anonymous
    • 5 years ago
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    1. 2x + 3 = 5 2. 2x + 5 = 3 3. 3x + 5 = 2 4. 3x + 2 = 5 5. 5x + 3 = 2 6. 5x + 2 = 3

  7. anonymous
    • 5 years ago
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    k. im with you so far

  8. anonymous
    • 5 years ago
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    ok cool, so now let's solve the cases

  9. anonymous
    • 5 years ago
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    1. 2x = 2. x = 1 (not negative)

  10. anonymous
    • 5 years ago
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    2. 2x = -2. x = -1 (negative)

  11. anonymous
    • 5 years ago
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    3. 3x = -3. x = -1. (negative)

  12. anonymous
    • 5 years ago
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    4. 3x = 3. x = 1. (positive)

  13. anonymous
    • 5 years ago
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    5. 5x = -1. x = -1/5. (negative)

  14. anonymous
    • 5 years ago
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    6. 5x = 1. x = 1/5 (positive)

  15. anonymous
    • 5 years ago
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    so out of the six possibilities, 3 are positive, 3 are negative

  16. anonymous
    • 5 years ago
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    so thats a 50 percent probabilty

  17. anonymous
    • 5 years ago
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    exactly

  18. anonymous
    • 5 years ago
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    and sometimes there's no systematic theorem or rule - it's just #outcomes/#possible outcomes

  19. anonymous
    • 5 years ago
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    right.

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spraguer (Moderator)
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