I'm just curious..

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I'm just curious..; How many PM's does DuffMan get a day?
Tweet Topic Started: Sep 13 2003, 07:22 PM (1,246 Views)
MiyamotoFan	Sep 13 2003, 07:22 PM Post #1
LB's GF (Brody = >:' D) Posts: 168 Group: Members Member #3,143 Joined: July 13, 2003	I have been wondering this for so long.



o7761146876	Sep 13 2003, 08:08 PM Post #2
Unregistered	well most likely...around 200 a day.. 4000 members 200 seems a reasionable answer
Unregistered


Eagle	Sep 13 2003, 08:12 PM Post #3
Member Posts: 290 Group: Members Member #5,099 Joined: September 9, 2003	i would guess most likely 800 a day.



o7761146876	Sep 13 2003, 08:15 PM Post #4
Unregistered	well the only way we gonna find out is...ask him
Unregistered


Jeff_Thomas	Sep 13 2003, 08:16 PM Post #5
Member Posts: 2,110 Group: Members Member #626 Joined: February 20, 2003	350 a week



o7761146876	Sep 13 2003, 08:27 PM Post #6
Unregistered	wow that's not alot that's like 50 a day not so bad
Unregistered


Codename	Sep 13 2003, 08:31 PM Post #7
Unregistered	800 a day is a lot.. I would say 200-400
Unregistered


Seth	Sep 13 2003, 10:05 PM Post #8
I has a pony Posts: 18,984 Group: Members Member #100 Joined: November 17, 2002	It's quite simple, really. Just utilize the formula P_M = (SQRT((2K^2)/(k-1))(2/(k+1))^^(k+1)/(k-1))(1-(P₂/P₁)^^(k-1)/k+((P₂-P)/P)(A^2) Where P_M is the number of PMs, K is the uncertainty factor, and P1 and P2 are the board population and the stupid people population of the world, respectively.



o7761146876	Sep 14 2003, 07:19 AM Post #9
Unregistered	Seth September 13, 09:05 PM It's quite simple, really. Just utilize the formula P_M = (SQRT((2K^2)/(k-1))(2/(k+1))^^(k+1)/(k-1))(1-(P₂/P₁)^^(k-1)/k+((P₂-P)/P)(A^2) Where P_M is the number of PMs, K is the uncertainty factor, and P1 and P2 are the board population and the stupid people population of the world, respectively. wow how did you find out the formuale
Unregistered


Crazy	Sep 14 2003, 07:22 AM Post #10
Member Posts: 956 Group: Members Member #4,539 Joined: August 24, 2003	Seth September 13, 09:05 PM It's quite simple, really. Just utilize the formula P_M = (SQRT((2K^2)/(k-1))(2/(k+1))^^(k+1)/(k-1))(1-(P₂/P₁)^^(k-1)/k+((P₂-P)/P)(A^2) Where P_M is the number of PMs, K is the uncertainty factor, and P1 and P2 are the board population and the stupid people population of the world, respectively. Damn..Wtf is that?



James	Sep 14 2003, 10:55 AM Post #11
Live to Dream Posts: 7,215 Group: Members Member #3,519 Joined: July 27, 2003	Seth, you forgot that there's only 30 days in this month - needs the * 2a + (n-1)d on the end



Crazy	Sep 14 2003, 10:56 AM Post #12
Member Posts: 956 Group: Members Member #4,539 Joined: August 24, 2003	:geek: And the geek awards go to... WWX and Seth! lol j/k



thunder	Sep 15 2003, 01:17 PM Post #13
Member Posts: 370 Group: Members Member #742 Joined: March 6, 2003	Seth September 13, 09:05 PM It's quite simple, really. Just utilize the formula P_M = (SQRT((2K^2)/(k-1))(2/(k+1))^^(k+1)/(k-1))(1-(P₂/P₁)^^(k-1)/k+((P₂-P)/P)(A^2) Where P_M is the number of PMs, K is the uncertainty factor, and P1 and P2 are the board population and the stupid people population of the world, respectively. hahaha oh that just brought up bad memories of math classes :blink:



pikapi	Sep 15 2003, 07:55 PM Post #14
I'm going to miss you... Posts: 478 Group: Members Member #838 Joined: March 16, 2003	Please no Math.. I suck at it



Creator	Sep 15 2003, 11:11 PM Post #15
iHeart IF Posts: 1,184 Group: Members Member #2,118 Joined: June 1, 2003	Seth... that comes out as -8.2xA9p+llama2 Hmm... very odd. Creator B)



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