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hba

Stats help needed

  • one year ago
  • one year ago

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  1. chihiroasleaf
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    what's your question?

    • one year ago
  2. hba
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    Good Question @chihiroasleaf

    • one year ago
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  3. hba
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    I have worked over this problem

    • one year ago
  4. uri
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    Faizan :o

    • one year ago
  5. hba
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    AM=4 x+3+2+0+8+y+3+4/8=4 x+y+20=32 x+y=12

    • one year ago
  6. hartnn
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    going good so far

    • one year ago
  7. hba
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    Now what i see is that 4 is given as the mode so x and y should also be 4 but that is not the case :/

    • one year ago
  8. hba
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    Looks like the question is wrong

    • one year ago
  9. hartnn
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    why x AND y ?? any one of them can be 4

    • one year ago
  10. hba
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    No but as its the mode so it should be the most recurring.....

    • one year ago
  11. hba
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    I mean to say that 3 occurs 2 times and it is not the mode.So 4 should occur 3 times to be the mode

    • one year ago
  12. hba
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    And if 3 is 2 times 4 is 2 times and then 8 would be 2 times,We have 3 modes here

    • one year ago
  13. hartnn
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    yes so what? we can select mode as any 1 of the 3, and we select here 4

    • one year ago
  14. hba
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    Okay got it,What should my step be now?

    • one year ago
  15. hba
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    What should i do after getting an eqn?

    • one year ago
  16. hartnn
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    take any one of 'x' or 'y' as 4

    • one year ago
  17. hba
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    Okay got it thanks a lot,but what should i state

    • one year ago
  18. hba
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    Why am i taking x or y as 4?

    • one year ago
  19. hba
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    I need to do some reasoning?

    • one year ago
  20. hartnn
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    because mode =4 if none of x or y is 4, then mode cannot be 4

    • one year ago
  21. hba
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    Got it bro thanks :D

    • one year ago
  22. hartnn
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    welcome :)

    • one year ago
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