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Howard-Wolowitz

  • one year ago

mean question

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  1. Howard-Wolowitz
    • one year ago
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  2. Michele_Laino
    • one year ago
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    using your data, we can write this: \[\Large \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 14\] where \( \large x_1,x_2,x_3,x_4,x_5\) are the requested positive numbers Now, if we choose \( \large x_3= 10 \), then we have: \[\Large {x_1},{x_2} < 10,\quad {x_4},{x_5} > 10\] by definition of median

  3. Howard-Wolowitz
    • one year ago
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    I see, but how would you simplify that

  4. Michele_Laino
    • one year ago
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    we can write this: \[\Large \begin{gathered} {x_1} + {x_2} + 10 + {x_4} + {x_5} = 70 \hfill \\ \hfill \\ \left( {{x_1} + {x_2}} \right) + \left( {{x_4} + {x_5}} \right) = 60 \hfill \\ \end{gathered} \] now I can choose \( \large x_1=5,x_2=9 \) so, what is \( \large x_3+x_4 =...?\)

  5. Howard-Wolowitz
    • one year ago
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    I dont know

  6. Howard-Wolowitz
    • one year ago
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    I would tihnk that you couldnt combine

  7. Michele_Laino
    • one year ago
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    if I substitute, I get this: \[\Large {x_4} + {x_5} = 60 - 9 - 5 = 46\]

  8. Michele_Laino
    • one year ago
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    so I can choose these values: \[\Large {x_4} = 22,\quad {x_5} = 24\]

  9. Howard-Wolowitz
    • one year ago
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    so is that what you would put to answer the first question?

  10. Michele_Laino
    • one year ago
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    yes!

  11. Michele_Laino
    • one year ago
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    here are the five positive numbers: \[\Large {x_1} = 5,\quad {x_2} = 9,\quad {x_3} = 10,\quad {x_4} = 22,\quad {x_5} = 24\]

  12. Howard-Wolowitz
    • one year ago
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    ok so i put those and thats it?

  13. Michele_Laino
    • one year ago
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    yes!

  14. Howard-Wolowitz
    • one year ago
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    alright cool

  15. Howard-Wolowitz
    • one year ago
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    what about what processi used

  16. Michele_Laino
    • one year ago
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    I have used the definition of \( \large median \) and the definition of \( \large mean \) of a distribution of data

  17. Howard-Wolowitz
    • one year ago
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    I see well ok that makes snese

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