Yahoo Answers: Answers and Comments for Can you find 6 numbers with a range of 7 a mean of 9 a median of 9 and a mode of 6 ? [Mathematics]
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From Lily
enGB
Sun, 17 Nov 2019 19:11:17 +0000
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Yahoo Answers: Answers and Comments for Can you find 6 numbers with a range of 7 a mean of 9 a median of 9 and a mode of 6 ? [Mathematics]
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From llaffer: 6 numbers with a range of 7.
That means the d...
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Sun, 17 Nov 2019 19:47:11 +0000
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.

From az_lender: try 6 6 6 11 12 13.
Right mean (9), right rang...
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Sun, 17 Nov 2019 19:26:17 +0000
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.

From Krishnamurthy: https://www.bing.com/search?q=6+numbers+with+a...
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https://uk.answers.yahoo.com/question/index?qid=20191117191117AAu9ax9
Mon, 18 Nov 2019 05:21:39 +0000
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