Question 988180
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If you want an average of *[tex \Large A] on *[tex \Large n] tests, then the sum of the scores of all *[tex \Large n] tests must be *[tex \Large An]


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

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