# The mean of the numbers.

• May 30th 2010, 02:33 PM
dnntau
The mean of the numbers.
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 =
• May 30th 2010, 02:37 PM
Mush
Quote:

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:

$\frac{a+b+c+d+e}{m} = \frac{a+b+c+d+e}{\frac{a+b+c+d+e}{5}} = (a+b+c+d+e) \times \frac{5}{a+b+c+d+e} = 5$
• May 30th 2010, 02:49 PM
dnntau
Thanks so much. (Rofl)
• May 30th 2010, 03:02 PM
pickslides
$\frac{a + b + c + d + e}{5} =m$

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?