Question 1161156
<font face="Times New Roman" size="+2">


Fill-in directly (Black numbers)


I & D Only: 60, Clue 1.  Region IV


I & R Only: 8, Clue 2.   Region V


None: 16, Clue 7.  Region VIII


Calculated (Magenta Numbers)


I but not D is any regions with and I but not a D.  So the total of regions I and V is 48 by clue 4.  However, we have region V as 8, so region I must be 48 minus 8, or 40.


I and R is any region with an I and an R regardless of the presence of a D, so by clue 6, the sum of regions V and VII must be 12, so VII must be 4 because V is 8.


From clue 5, the sum of VI and VII is 20, so knowing VII is 4, gives VI = 16.


Clue 3 says region III and VI must add up to 28; hence region III must be 8.


Since this is an analysis of 200 responses, the sum of all the regions on the graph must be 200.  Since we have all but one of the regions filled in, the remaining region, namely II, is 200 minus the sum of all the others.


Then the answer to the question posed is the sum of regions I, II, and III.


You can do the rest of the arithmetic yourself.


 *[illustration Venn.jpg]

								
								
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>