Expert: Sherman D. - 3/20/2006

Question
Hello:

I want to thank you for the reply, but I think that you reversed the process.  The 2/3 and 1/3 are not known. I do not believe that your calculation will not work.
I want to know how to determine these fractions, 2/3 and 1/3 in my example. The calculation would be something like this: (? X 5) + (? X 2) = 4, where the first ? represents 2/3 and the second ?  represents 1/3.  I know that the simple average is (1/2 X 5) + (1/2 X 2) = 3.5.

I thank you for your follow-up reply.
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Followup To
Question -
Hello:

The simple average for 5 and 2 is 3.5, (1/2 X 5) + (1/2 X 2) = 5/2 + 2/2 = 7/2 = 3.5. However, if 4 is preferred for the average, a weighted average is used as follows: (2/3 X 5) + (1/3 X 2) = 10/3 + 2/3 = 12/3 = 4.

How are these two fractions, 2/3 and 1/3, determined for making the average 4 instead of 3.5 for 5 and 2?
If possible, use a simple procedure or calculation.

I thank you for your reply.
Answer -
((2/3)x) + ((1/3)y) = 4

((2x)/3) + (y/3) = 4

multiply everything by 3

2x + y = 12

y = -2x + 12

Here are some numbers

0,12
1,10
2,8
3,6
4,4
6,0

"x" or "y" can't be negative, so therefore you can't use anything below 0 or anything higher than 6.

by the way, since you wanted to find other numbers besides 5 and 2, i just replaced 5 with "x" and 2 with "y" to find the other numbers.

Answer
same process

(x * 5) + (y * 2) = 4

5x + 2y = 4

2y = -5x + 4

y = (-5/2)x + 2

Any number between 0 and (4/5) for x are the only values that can be used.

0, 2
to
(4/5), 0

Other than that, thats about all i can tell you.

Averages are found like this (a + b + c + d + ... nth)/z where as z is the number of values

(a/z) + (b/z) + (c/z) + (d/z) + ... ((nth)/z)

As you saw above that (2/3 * 5) and (1/3 * 2) became 10/3 + 2/3, thats the same thing as saying "a" = 10, "b" = 2, and "z" = 3