1. **State the problem:** We are given three averages: - $x$ is the average of $m$ and 9. - $y$ is the average of $2m$ and 15. - $z$ is the average of $3m$ and 18. We need to find the average of $x$, $y$, and $z$ in terms of $m$. 2. **Recall the formula for average (arithmetic mean):** $$\text{Average of two numbers } a \text{ and } b = \frac{a+b}{2}$$ 3. **Calculate each average:** - $x = \frac{m + 9}{2}$ - $y = \frac{2m + 15}{2}$ - $z = \frac{3m + 18}{2}$ 4. **Find the average of $x$, $y$, and $z$:** $$\text{Average} = \frac{x + y + z}{3} = \frac{\frac{m + 9}{2} + \frac{2m + 15}{2} + \frac{3m + 18}{2}}{3}$$ 5. **Combine the fractions inside the numerator:** $$= \frac{\frac{(m + 9) + (2m + 15) + (3m + 18)}{2}}{3} = \frac{\frac{6m + 42}{2}}{3}$$ 6. **Simplify the numerator:** $$= \frac{6m + 42}{2} \times \frac{1}{3} = \frac{6m + 42}{6}$$ 7. **Simplify the fraction:** $$= \frac{6m}{6} + \frac{42}{6} = m + 7$$ **Final answer:** The average of $x$, $y$, and $z$ in terms of $m$ is $$m + 7$$