Question 981297
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If you have a *[tex \Large \mu_0] average for *[tex \Large n] scores, then you have a total of *[tex \Large n\mu_0] points.  In order to have a *[tex \Large \mu_1] average for *[tex \Large n\ +\ 1] scores, your total points for those *[tex \Large n\ +\ 1] scores must be *[tex \Large (n\ +\ 1)\mu_1].  The score needed for the last test is then *[tex \Large (n\ +\ 1)\mu_1\ -\ n\mu_0]


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

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