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hba

  • 3 years ago

Stats help needed

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  1. chihiroasleaf
    • 3 years ago
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    what's your question?

  2. hba
    • 3 years ago
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    Good Question @chihiroasleaf

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  3. hba
    • 3 years ago
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    I have worked over this problem

  4. uri
    • 3 years ago
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    Faizan :o

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

  6. hartnn
    • 3 years ago
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    going good so far

  7. hba
    • 3 years ago
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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 :/

  8. hba
    • 3 years ago
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    Looks like the question is wrong

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

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

  11. hba
    • 3 years ago
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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

  12. hba
    • 3 years ago
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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

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

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

  15. hba
    • 3 years ago
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    What should i do after getting an eqn?

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

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

  18. hba
    • 3 years ago
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    Why am i taking x or y as 4?

  19. hba
    • 3 years ago
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    I need to do some reasoning?

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

  21. hba
    • 3 years ago
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    Got it bro thanks :D

  22. hartnn
    • 3 years ago
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    welcome :)

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