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To: paintriot

How many election cycles would there have to have been to come up with 55% participation?

Smart FReepers, anyone?

“Math is hard”


34 posted on 01/31/2016 2:52:54 PM PST by digger48
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To: digger48
Nine, of course.

5 / 9 = 0.55555 . . .

64 posted on 01/31/2016 4:16:28 PM PST by goldbux (CDO / I may have Obsessive Compulsive Disorder, but at least I put the letters in correct sequence.)
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To: digger48; goldbux
How many election cycles would there have to have been to come up with 55% participation?

Congrats on asking the $64,000 question: the answer to which demonstrates that the numbers are almost certainly made up. The answer is 20, if you account for the alternative scores of 65% and 75% at post 21.

11/20=55%

13/20=65%

15/20=75%

BUT according to the linked article, Peterson, who got a 55% score, "moved to Iowa in 2009." How likely is it that 20 elections have occurred in Peterson's precinct since 2009? I suppose it is possible with school levies, etc., but I deem the likelihood very small. This assessment is reinforced by another poster on this thread, who says that caucus attendance is not recorded by the state. If correct, caucuses presumably would not be counted.

73 posted on 01/31/2016 6:23:17 PM PST by matt1234 (Note to GOPe lurkers: I and thousands like me will NEVER vote for Jeb Bush)
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