Problem Analysis:

• We are given nine numbers, $x_1, x_2, ..., x_9$, sorted in ascending order: $x_1 \le x_2 \le ... \le x_9$.
• Their average is $m = \frac{x_1 + x_2 + ... + x_9}{9}$.
• We are told $m$ is strictly greater than the first eight numbers: $m > x_i$ for $i = 1, 2, ..., 8$.
• We need to evaluate two statements involving averages.

Evaluating Average $(x_1, x_2, ..., x_9, m)$

This average includes the original nine numbers plus the average $m$. There are 10 numbers in total.

1. Sum of the numbers: The sum is $(x_1 + x_2 + ... + x_9) + m$.
2. Substitute using the definition of $m$: From $m = \frac{x_1 + ... + x_9}{9}$, we know $x_1 + ... + x_9 = 9m$.
3. Total Sum: The sum becomes $9m + m = 10m$.
4. Calculate the Average: The average is $\frac{10m}{10} = m$.

Therefore, Average $(x_1, x_2, ..., x_9, m) = m$. This matches the first condition in Option C.

Evaluating Average $(x_2, x_3, ..., x_9)$

This average considers the numbers from $x_2$ to $x_9$. There are 8 numbers in total.

1. Sum of these 8 numbers: $S = x_2 + x_3 + ... + x_9$.
2. Relate to $m$: We know $9m = x_1 + x_2 + ... + x_9$. So, $S = 9m - x_1$.
3. Calculate the Average: The average is $A = \frac{S}{8} = \frac{9m - x_1}{8}$.
4. Compare $A$ with $m$: We need to determine if $A > m$, $A < m$, or $A = m$. Let's look at the difference $A - m$: $A - m = \frac{9m - x_1}{8} - m$ $A - m = \frac{9m - x_1 - 8m}{8}$ $A - m = \frac{m - x_1}{8}$
5. Use the given condition: The problem states $m > x_1$ (since $m$ is strictly greater than all the first eight numbers).
6. Conclusion: Because $m > x_1$, the term $(m - x_1)$ is positive. Therefore, $\frac{m - x_1}{8} > 0$, which means $A - m > 0$, or $A > m$.

Thus, Average $(x_2, x_3, ..., x_9) > m$. This matches the second condition in Option C.

Final Conclusion

Based on the calculations:

• Average $(x_1, x_2, ..., x_9, m) = m$
• Average $(x_2, x_3, ..., x_9) > m$

Both conditions are met by Option C.