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January 25th, 2019, 07:04 PM  #1 
Newbie Joined: Jan 2019 From: Canada Posts: 3 Thanks: 0  Percentage question and solving given average
Hi I'm new here. I have a question about solving for an average number. It might be pretty basic but I'm just too tired to think tonight. I am trying to solve for what two proportions make up an average of 44% given that a occurs 73% of the time and another thing occurs 27% of the time. I am trying to work it out but super tired. I'm thinking one number is something like 35% and the other is something like 48%. I appreciate any help. 
January 25th, 2019, 07:11 PM  #2 
Newbie Joined: Jan 2019 From: Canada Posts: 3 Thanks: 0 
Bah I'm silly I guess the weights don't say anything if I don't have some proportion. It could just be 44% for both spots, but on average the O number is about 3/4 to 4/5ths that of BB call #. [((3a x 0.27)/4)+(a x 0.73)]=44 Is this the correct formula? No clue what I'm doing to be honest. 4x(3a x 0.27)/4 + (a x 0.73)/0.73) = 44 12a x 1.08+1a = 44 13a=44/1.08 13a=40.74 a= 3.13 12 x a or 3.13= 37.56% for first number? Seems reasonable. 44 x 1.08 = 47.52% for 2nd number. I'm pretty sure this is all wrong, but I'm just trying anyway. (47.5x0.73)+(37.5x0.27)=44 34.69+10.13=44.82 damn close Last edited by skipjack; January 26th, 2019 at 05:58 AM. 
January 26th, 2019, 05:56 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,484 Thanks: 2041 
As 44  27 = 17 and 73  44 = 29, using (m  p)q + (q  m)p ≡ m(q  p), one gets (17/46)*73 + (29/46)*27 = 44. Last edited by skipjack; January 26th, 2019 at 07:27 PM. 
January 26th, 2019, 08:42 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,342 Thanks: 1024 
In case useful for you: Q = quantity, P = percentage ....Q......P ....a.....u% ....b.....v%  ..a+b...w% w = (au + bv) / (a + b) 
January 26th, 2019, 11:08 AM  #5 
Newbie Joined: Jan 2019 From: Canada Posts: 3 Thanks: 0 
Ty


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