Question 983292
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If the mean of *[tex \Large n] numbers is *[tex \Large \mu], then the sum of the numbers is *[tex \Large n\mu]


If *[tex \Large a] is added to the group, the sum of the numbers is *[tex \Large n\mu\ +\ a]


The mean of the *[tex \Large n\ +\ 1] numbers is then *[tex \Large \frac{n\mu\ +\ a}{n\ +\ 1}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \