answer:To analyze each option systematically: **Option A**: We need to compare the average of x_{2}, x_{3}, x_{4}, x_{5} with the average of all six values x_{1}, x_{2}, cdots, x_{6}. The average of four values is frac{x_{2} + x_{3} + x_{4} + x_{5}}{4}, and the average of all six values is frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6}. Without specific values, we cannot guarantee these two averages are equal because x_{1} and x_{6} can significantly influence the overall average. Therefore, **Option A is incorrect**. **Option B**: The median of x_{2}, x_{3}, x_{4}, x_{5} is the average of x_{3} and x_{4}, which is frac{x_{3} + x_{4}}{2}. For the original set x_{1}, x_{2}, cdots, x_{6}, since there are an even number of values, the median is also the average of the middle two values, which are again x_{3} and x_{4}. Thus, the median of both sets is frac{x_{3} + x_{4}}{2}. This means **Option B is correct**. **Option C**: To examine this, consider a specific example where x_{1} = 0, x_{2} = 1, x_{3} = 2, x_{4} = 8, x_{5} = 9, and x_{6} = 10. The average of all six values is 5, and the average of x_{2}, x_{3}, x_{4}, x_{5} is also 5. Calculating the variance for both sets, we find that the variance of the six values is frac{50}{3}, and the variance of the four values is frac{25}{2}. Since variance is directly related to standard deviation, and frac{50}{3} > frac{25}{2}, it shows that the standard deviation of the full set is greater than that of the subset excluding the minimum and maximum. Thus, **Option C is incorrect**. **Option D**: The range of a set is the difference between its maximum and minimum values. For x_{2}, x_{3}, x_{4}, x_{5}, the range is x_{5} - x_{2}. For the full set x_{1}, x_{2}, cdots, x_{6}, the range is x_{6} - x_{1}. Since x_{6} is greater than x_{5} and x_{2} is greater than x_{1}, it follows that the range of the full set is greater than the range of the subset, making **Option D correct**. Therefore, the correct options are boxed{B text{ and } D}.

answer:To determine the maximum element in the set ( A = {sqrt[n]{n} mid n in mathbf{N} text { and } 1 leqslant n leqslant 2020 } ), we will analyze the function ( f(x) = frac{ln x}{x} ). 1. Consider the function ( f(x) = frac{ln x}{x} ). 2. Determine the critical point of ( f(x) ) by finding its first derivative and setting it to zero: [ f'(x) = frac{d}{dx} left(frac{ln x}{x}right) = frac{1 - ln x}{x^2}. ] 3. Set ( f'(x) = 0 ) to find the critical points: [ frac{1 - ln x}{x^2} = 0 implies 1 - ln x = 0 implies ln x = 1 implies x = e. ] 4. Analyze the behavior of ( f(x) ) on the intervals ( (0, e) ) and ( (e, +infty) ): - On the interval ( (0, e) ): [ f'(x) > 0 implies f(x) = frac{ln x}{x} text{ is increasing}. ] - On the interval ( (e, +infty) ): [ f'(x) < 0 implies f(x) = frac{ln x}{x} text{ is decreasing}. ] 5. Therefore, ( f(x) ) attains its maximum at ( x = e ). Since ( e ) is approximately 2.718, for integers ( n ), we analyze the values around ( n = 3 ) and ( n = 2 ): - Compute ( frac{ln n}{n} ) for ( n = 3 ): [ f(3) = frac{ln 3}{3}. ] - Compute ( frac{ln n}{n} ) for ( n = 4 ): [ f(4) = frac{ln 4}{4}. ] 6. Compare ( f(3) ) and ( f(4) ): [ frac{ln 3}{3} > frac{ln 4}{4}. ] 7. Generalize by induction for other values ( n ): [ frac{ln 3}{3} > frac{ln 4}{4} > dots > frac{ln 2020}{2020}. ] 8. Transform back using the exponential function: [ ln sqrt[3]{3} > ln sqrt[4]{4} > dots > ln sqrt[2020]{2020} implies sqrt[3]{3} > sqrt[4]{4} > dots > sqrt[2020]{2020}. ] 9. Additionally, verify that: [ sqrt[3]{3} > sqrt{2} > 1. ] Conclusion: The maximum element of the set ( A ) is ( sqrt[3]{3} ). [ boxed{sqrt[3]{3}} ]

answer:1. **Determine the nature of the problem:** - We are tasked to color a connected graph where each node represents a camp (营地) and each edge indicates adjacency. - We must ensure that adjacent nodes have different colors. - The goal is to find the minimum number of colors required to achieve this. 2. **Analyze the given graph structure:** - According to the problem, there are 9 camps. - The problem states that 3 points in the middle form a triangle, which is pivotal as it indicates a 3-cycle (a connected component with 3 nodes all mutually adjacent). 3. **Identify the chromatic number of a cycle:** - A cycle of 3 nodes (triangle) requires at least 3 colors to ensure that no two adjacent nodes share the same color. - This is known from graph coloring theory, where the chromatic number of a complete graph ( K_3 ) is 3. 4. **Including additional nodes and extensions:** - We need to ensure the camps outside the triangle are also colored such that adjacent camps have different colors. - The triangle is the most constrained part; thus, we start by assigning 3 different colors to it. 5. **Propagate colors to the remaining camps:** - Any additional node connected to one of these triangle nodes can use any of the 3 colors, ensuring it is different from its adjacent node(s). 6. **Count the minimum colors required:** - Since we most centrally have a complete subgraph ( K_3 ) within our larger graph, we will need at least 3 colors. - Even if extending the graph with more nodes, trivially connecting them in such a way always keeps the requirement at a minimum of 3 distinct colors for non-adjacent nodes. 7. **Conclusion:** - Thus, the minimum number of colors required is ( boxed{3} ).

answer:The farmer had 127 apples originally and now has 39 apples left. To find out how many apples he gave to his neighbor, we subtract the number of apples he has left from the original number of apples. 127 apples (original) - 39 apples (left) = 88 apples (given to the neighbor) The farmer gave boxed{88} apples to his neighbor.