The mean of the numbers.

**dnntau** · May 30th 2010, 01:33 PM

How would I find...
If the mean of the numbers a, b, c, d, and e is m and m is not equal to 0, then a + b + c + d + e/ m =

**Mush** · May 30th 2010, 01:37 PM

Originally Posted by dnntau

How would I find...
If the mean of the numbers a, b, c, d, and e is m and m is not equal to 0, then a + b + c + d + e/ m =

Remember that the mean is given by:

$m = \frac{a+b+c+d+e}{5}$

Hence:

$\frac{a+b+c+d+e}{m} = \frac{a+b+c+d+e}{\frac{a+b+c+d+e}{5}} = (a+b+c+d+e) \times \frac{5}{a+b+c+d+e} = 5$

**dnntau** · May 30th 2010, 01:49 PM

Thanks so much.

**pickslides** · May 30th 2010, 02:02 PM

$\frac{a + b + c + d + e}{5} =m$

Find

$\frac{a + b + c + d + e}{ m}$

$\frac{a + b + c + d + e}{\frac{a + b + c + d + e}{5} }$

$(a + b + c + d + e) \times\frac{5}{a + b + c + d + e}$

Then what?

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