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January 25th, 2019, 08:04 PM   #1
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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.
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January 25th, 2019, 08:11 PM   #2
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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 06:58 AM.
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January 26th, 2019, 06:56 AM   #3
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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 08:27 PM.
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January 26th, 2019, 09:42 AM   #4
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In case useful for you:

Q = quantity, P = percentage

....Q......P
....a.....u%
....b.....v%
------------
..a+b...w%

w = (au + bv) / (a + b)
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January 26th, 2019, 12:08 PM   #5
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Ty
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