Answer:
C
Step-by-step explanation:
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,
x=(m+9)/2
y=(2m+15)/2
z=(3m+18)/2
the average of x, y, and z
=(x+y+z)/3
=(m+9+2m+15+3m+18)/3
=(6m+42)/3
=2m+14
ans is C
Answer: C). 2m+14
Step-by-step explanation:
[tex]\left[\begin{array}{ccc}x=(m+9)/2\\y=(2m+15)/2\\z=(3m+18)/2\end{array}\right][/tex]
The sequence above would be the sequence of x,y and z. So them we would divide all x,y and z and divide them by 3 it's self. Then we do the following:
[tex]\boxed{m*(x,y,z)}[/tex]
Then after we multiple all x,y and z, we then get 42. We then do (42+6m) ÷ 3...
[tex]\boxed{\boxed{Answer: \bf2m+14}}}[/tex]
