<PRE><B>Question 770552</B>
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If the mean for 2 years was some number, then the sum of the attendances for the two years was 2 times the mean, so if you let *[tex \LARGE x] represent the attendance in 2006, you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ +\ 120\ =\ 90530]


Solve for *[tex \LARGE x].  Then calculate *[tex \LARGE x\ +\ 120]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
&lt;div style=&quot;text-align:center&quot;&gt;&lt;a href=&quot;http://outcampaign.org/&quot; target=&quot;_blank&quot;&gt;&lt;img src=&quot;http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png&quot; border=&quot;0&quot; alt=&quot;The Out Campaign: Scarlet Letter of Atheism&quot; width=&quot;143&quot; height=&quot;122&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
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