Question

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

Answer 1

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 2

Answer: C). 2m+14

Step-by-step explanation:

$\left[\begin{array}{ccc}x=(m+9)/2\\y=(2m+15)/2\\z=(3m+18)/2\end{array}\right]$

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:

$\boxed{m*(x,y,z)}$

Then after we multiple all x,y and z, we then get 42. We then do (42+6m) ÷ 3...

$\boxed{\boxed{Answer: \bf2m+14}}}$