Question 1129668
<font face="Times New Roman" size="+2">


If the average of a set of *[tex \Large n] numbers is 23, then the sum of those *[tex \Large n] numbers must be *[tex \Large 23n].  If you add -35 to that sum and divide by *[tex \Large n\ +\ 1] you get the new average of 21, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{23n\ -\ 35}{n\ +\ 1}\ =\ 21]


Solve for *[tex \Large n]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>