Can you find 6 numbers with a range of 7 a mean of 9 a median of 9 and a mode of 6 ?

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  • 3 weeks ago

    6 numbers with a range of 7.

    That means the difference of the max and min is 7.

    So  let's come up with variables for your 6 numbers, then an expression for the last one:

    a, b, c, d, e, f

    We know that f is 7 more than a, so:

    f = a + 7

    Leaving us with these 6 numbers:

    a, b, c, d, e, (a + 7)

    The mean and median are both 9.

    The median of an even number of values is the mean of the middle to (c and d).  So:

    (c + d) / 2 = 9

    c + d = 18

    And we know the mean of all 6 values is 9, so:

    (a + b + c + d + e + a + 7) / 6 = 9

    2a + b + c + d + e + 7 = 54

    2a + b + c + d + e = 47

    We know that c + d = 18, so:

    2a + b + (c + d) + e = 472a + b + 18 + e = 472a + b + e = 29

    We know the mode is 6.  So two of these three numbers has to be a 6.  Since "e" has to be greater than 9, that leaves "a" and "b" to be 6.  So now we can solve for "e":

    2(6) + 6 + e = 2912 + 6 + e = 2918 + e = 29e = 11Going back to our numbers we now have:

    6, 6, c, d, 11, (a + 7)

    We know "a" so we know (a + 7):

    6, 6, c, d, 11, 13The only thing left now is that we know the mean of c and d must be 9.  So back to this:

    (c + d) / 2 = 9

    c + d = 18

    So like the last one, we have a few possible answers.  If we keep these to only integers and know the must be 18 and the two values must be greater than 6 and less than 11 and not equal to each other, there is only one possible set this can be:

    6, 6, 8, 10, 11, 13

    If c was 6 (which would still make 6 a mode), d would have to be 12, which makes the median equal to 10, which breaks that rule.

    If c was 7, d would have to be 11, which doesn't make 6 the only mode.

    If c was 9, d would be 9, which also doesn't ake 6 the only mode.

    Those are why those values are excluded to leave the only set left to be correct.

  • 3 weeks ago

    try 6 6 6 11 12 13.

    Right mean (9), right range (7), right mode (6),

    wrong median (8.5).

    try 6 6 x y z 13, requiring that

    x + y = 18, x + y + z = 29, and z > y.

    Then z must be 11, so x and y could be 8 and 10.

    6 6 8 10 11 13.

    Right mean, right median, right mode, right range.

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