+0  
 
+1
35
4
avatar+203 

If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m?

 

         A) m+6
      B) m+7
   C) 2m+14
D) 3m+21

 Feb 26, 2019
 #1
avatar+17275 
+4

I come up with m+7

 

(x+y+z)/3 is the average....sub in the values given for x y and z

 

 

([9 +m   +2m+15  + 3m+18] /2   )/3  = ( (6m + 42)/2) / 3 = (3m+21)/3  = m+7

 Feb 26, 2019
 #2
avatar+98091 
+3

Just want to try this one....EPs answer might be better

 

We have that

 

[ m + 9 ] / 2 = x         [2m + 15] / 2 = y       [ 3m + 18 ] / 2 = z

 

So  the average of x,  y , z is

 

[ ( m + 9)  + (2m + 15) + (3m + 18) ] /  ( 2 * 3)  =

 

[ 6m + 42] / 6  =

 

m + 7

 

 

cool cool cool

 Feb 26, 2019
 #3
avatar+98091 
+2

We got the same result.....must be correct.....LOL!!!

 

 

cool cool cool

 Feb 26, 2019
 #4
avatar+203 
+2

yeah right haha laugh

 Feb 26, 2019

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