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Let's break this problem down step by step:

### Step 1: Define the values of \( x \), \( y \), and \( z \).

- \( x \) 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:
  \[
  z = \frac{3m + 18}{2}
  \]

### Step 2: Find the average of \( x \), \( y \), and \( z \).

To find the average of \( x \), \( y \), and \( z \), we add them together and divide by 3:
\[
\text{Average of } x, y, z = \frac{x + y + z}{3}
\]

Substitute the values of \( x \), \( y \), and \( z \) into this expression:
\[
\frac{x + y + z}{3} = \frac{\left( \frac{m + 9}{2} \right) + \left( \frac{2m + 15}{2} \right) + \left( \frac{3m + 18}{2} \right)}{3}
\]

### Step 3: Simplify the expression.

Combine all the terms in the numerator:
\[
x + y + z = \frac{m + 9}{2} + \frac{2m + 15}{2} + \frac{3m + 18}{2}
\]
Since they have the same denominator, we can add the numerators:
\[
x + y + z = \frac{(m + 9) + (2m + 15) + (3m + 18)}{2}
\]
Simplify the terms inside the parentheses:
\[
x + y + z = \frac{6m + 42}{2}
\]
Now, simplify further:
\[
x + y + z = 3m + 21
\]

### Step 4: Find the final average.
Now, divide by 3 to find the average of \( x \), \( y \), and \( z \):
\[
\frac{x + y + z}{3} = \frac{3m + 21}{3} = m + 7
\]

### Final Answer:
The average of \( x \), \( y \), and \( z \) is \( m + 7 \), which corresponds to option **B**.

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Do you have any questions or would you like further details?

### 5 Related Questions:
1. What is the difference between the arithmetic mean and the geometric mean?
2. How do you find the weighted average if different values have different weights?
3. How does the average change if one of the numbers is doubled?
4. How would this problem change if one of the numbers involved was negative?
5. How can we apply this process to find the average of more complex functions?

### Tip:
Always simplify expressions step by step, combining like terms and using parentheses where necessary to avoid errors!

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?

Solve a problem involving averages and algebra by finding the arithmetic mean of three variables x, y, and z in terms of m. This step-by-step guide helps students in Grades 9-12 understand key algebraic concepts and arithmetic mean calculations.

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