Question 992290
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If you have *[tex \Large n] scores and need to know the score on the *[tex \Large n\ +\ 1]th assignment in order to have an average of *[tex \Large \mu], then multiply *[tex \Large n\ +\ 1] times *[tex \Large \mu] and subtract the sum of the first *[tex \Large n] scores.


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

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