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:

B

Explanation:

Let's convert these written languages into mathematical expressions:

"x is the average of m and 9": x = $\frac{m +9}{2}$

"y is the average of 2m and 15": y = $\frac{2m +15}{2}$

"z is the average of 3m and 18": z = $\frac{3m +18}{2}$

Now, we want to find the value of the average of x, y, and z in terms of m. The average of x, y, and z can be written as: $\frac{x+y+z}{3}$. Let's substitute each of the expressions we have above in for x, y, and z:

$\frac{\frac{m +9}{2}+\frac{2m +15}{2}+\frac{3m +18}{2}}{3}=\frac{(m+9)+(2m+15)+(3m+18)}{6} =\frac{6m+42}{6} =m+7$

The answer is B.

Hope this helps!

Answer 2

Answer:

B.) M + 7

Explanation:

Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2, the equations X =  m+9/ 2 , Y =  2m+15/2 , Z =  3m+18 /2  are true. The average of x, y, and z is given by  x+y+z /3 . Substituting the expressions in m for each variable (x, y, z) gives:

[ m+9 /2 + 2m+15/ 2 + 3m+18/2 ]  / 3

This fraction can be simplified to m+7.