Question 330550
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Remember how you calculate an average.  You add the numbers and divide by the number of numbers.  So if the average of 3 scores was 88, then the sum of the three scores must be 3 times 88 or 264.


Let *[tex \Large x] represent the score on the first test.  Then the score on the second test must be *[tex \Large x\ +\ 10], and the score on the third test must be *[tex \Large x\ +\ 10\ +\ 4\ =\ x\ +\ 14].


Add 'em up:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ +\ 10\ +\ x\ +\ 14\ =\ 264]


Solve for *[tex \Large x], then add 10 and then add 4 more.


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