answer:1. **Determine total number of vertices**: A dodecahedron has 20 vertices. 2. **Total ways to choose 2 vertices**: The number of ways to choose 2 vertices from 20 is given by the combination formula binom{20}{2} = frac{20 times 19}{2} = 190. 3. **Vertices connected by an edge**: Each vertex of a dodecahedron has 3 edges connected to other vertices. Thus, from any one vertex, you can choose 3 other vertices that directly share an edge with it. 4. **Total favorable outcomes**: Since each of the 20 vertices connects to 3 other vertices, you initially get 20 times 3 = 60. However, this count considers each edge twice (once from each endpoint), so the exact number of unique edges is frac{60}{2} = 30. 5. **Probability calculation**: The probability that two randomly chosen vertices share an edge is the ratio of favorable outcomes to the total outcomes, which is frac{30}{190} = frac{3}{19}. Thus, the probability that two randomly chosen vertices from a dodecahedron are endpoints of an edge is boxed{frac{3}{19}}.

answer:In the new pie chart scenario, we are interested in the counties that have a population between 20,000 and 150,000 residents. According to the given information: - The segment representing counties with 20,000 to 150,000 residents accounts for 45% of the total. Using the information provided: [ text{Percentage of counties with 20,000 to 150,000 residents} = 45% ] Therefore, the final answer is: [ boxed{45%} ]

answer:(1) Given an arithmetic sequence {a_n}, we have a_1 + a_3 + a_5 = 6 and a_1 cdot a_3 cdot a_5 = 0. Since a_3 = a_1 + 2d, the equations can be rewritten as: [ begin{cases} (a_3 - 2d) + a_3 + (a_3 + 2d) = 6, (a_3 - 2d) cdot a_3 cdot (a_3 + 2d) = 0. end{cases} ] From the above system, we can deduce that a_3 = 2 and the common difference d = -1. Hence, the general term for the sequence is: [ a_n = a_3 + (n-3)d = 2 + (n-3)(-1) = 5 - n. ] The final formula for the general term is boxed{a_n = 5 - n}. (2) Now for b_n = frac{1}{n(a_n - 6)}, we substitute the expression for a_n: [ b_n = frac{1}{n(5 - n - 6)} = frac{1}{n(1 - n)} = - frac{1}{n (n - 1)}. ] By partial fraction decomposition, we can express b_n as: [ b_n = frac{1}{n} - frac{1}{n - 1}. ] This allows us to write the sum S_n as a telescoping series: [ S_n = left( frac{1}{2} - frac{1}{1} right) + left( frac{1}{3} - frac{1}{2} right) + cdots + left( frac{1}{n} - frac{1}{n - 1} right) + left( frac{1}{n + 1} - frac{1}{n} right), ] which sums up to boxed{S_n = - frac{n}{n + 1}}.

answer:# Problem: On the coordinate plane, identify the sets of points whose coordinates ((x, y)) satisfy the following conditions: 1) (x^{2} y + y^{3} = 2x^{2} + 2y^{2}) 2) (xy + 1 = x + y) **Part 1:** 1. Starting with the equation ( x^2 y + y^3 = 2x^2 + 2y^2 ). 2. We will rearrange the equation to move all terms to one side: [ x^2 y + y^3 - 2x^2 - 2y^2 = 0. ] 3. Notice that we can factor ( x^2 ) and ( y^2 ) from the corresponding terms: [ x^2 y - 2x^2 + y^3 - 2y^2 = 0, ] which gives [ x^2 (y - 2) + y^2 (y - 2) = 0. ] 4. Factor out ( y - 2 ): [ (x^2 + y^2)(y - 2) = 0. ] 5. This product is zero if ( x^2 + y^2 = 0 ) or ( y - 2 = 0 ). 6. Since ( x^2 + y^2 = 0 ) implies ( x = 0 ) and ( y = 0 ) (since ( x^2 ) and ( y^2 ) are non-negative and their sum zero means both must be zero): [ text{This gives the point } (0, 0). ] 7. Alternatively, the equation ( y - 2 = 0 ) gives: [ y = 2. ] 8. Hence, the solution set for part 1 consists of the point ( (0, 0) ) and the line ( y = 2 ). **Part 2:** 1. The given equation is ( xy + 1 = x + y ). 2. Rearrange the equation: [ xy + 1 - x - y = 0, ] which simplifies to: [ xy - x - y + 1 = 0. ] 3. Notice that we can factor this equation as: [ (x-1)(y-1) = 0. ] 4. This equation is satisfied if ( x-1 = 0 ) or ( y-1 = 0 ), which implies: [ x = 1 quad text{or} quad y = 1. ] 5. Therefore, the solution set for part 2 consists of the lines ( x = 1 ) and ( y = 1 ). # Conclusion: The combined solution sets are: - For part 1: the line ( y = 2 ) and the point ( (0, 0) ). - For part 2: the lines ( x = 1 ) and ( y = 1 ). [ boxed{(0, 0), y = 2, x = 1, y = 1} ]