<PRE><B>Question 587371</B>
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If you must have an average of *[tex \LARGE A] on *[tex \LARGE n] tests, then the sum of the scores on *[tex \LARGE n] tests must be *[tex \LARGE nA].  Then, if you already know the scores on *[tex \LARGE n\ -\ 1] of the tests, add those scores and subtract from *[tex \LARGE nA].  The result will be the score necessary to achieve the desired average.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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