answer:(I) Since the ellipse (C): frac{x^{2}}{a^{2}} + frac{y^{2}}{b^{2}} = 1 (a > b > 0) has an eccentricity of frac{sqrt{3}}{2} and a right focus at F(3, 0), we have c = 3 and frac{c}{a} = frac{sqrt{3}}{2}. Solving for a and b, we get a = 2sqrt{3} and b = sqrt{3}. Thus, the standard equation of ellipse (C) is frac{x^{2}}{12} + frac{y^{2}}{3} = 1. (II) Let N(4, m), A(x_1, y_1), and B(x_2, y_2), with M(x_0, y_0) as the midpoint of AB. The slope of FN is m. Since F(3, 0), we can write the equation of line AB as x = -my + 3. Substituting this into the ellipse equation, we obtain (m^2 + 4)y^2 - 6my - 3 = 0. Solving for y_1 and y_2, we find Mleft(frac{12}{m^2 + 4}, frac{3m}{m^2 + 4}right). The slope of line OM is k_{OM} = frac{m}{4}. Since k_{ON} = frac{m}{4} as well, k_{OM} = k_{ON}, which implies that points O, M, and N are collinear. (III) From (II), we know that Mleft(frac{12}{m^2 + 4}, frac{3m}{m^2 + 4}right), N(4, m), and points O, M, and N are collinear. Given that 2|OM| = |MN|, we have 2 cdot frac{12}{m^2 + 4} = 4 - frac{12}{m^2 + 4}. Solving for m, we get m = pm sqrt{5}. Thus, the equation of line (l) is x = pmsqrt{5}y + 3 or boxed{x pm sqrt{5}y - 3 = 0}.

answer:**Analysis** To minimize the total waiting time (including the time to fill the buckets), the buckets should be filled starting with the one that takes the least amount of time, in the order of 4, 5, 6, 8, 10. - The first person takes 4 minutes to fill their bucket: at this time, 4 people are waiting, so the total waiting time is 4 times 4 = 16 minutes. - The second person takes 5 minutes to fill their bucket: at this time, 3 people are waiting, so the total waiting time is 5 times 3 = 15 minutes. - The third person takes 6 minutes to fill their bucket: at this time, 2 people are waiting, so the total waiting time is 6 times 2 = 12 minutes. - The fourth person takes 8 minutes to fill their bucket: at this time, 1 person is waiting, so the total waiting time is 8 minutes. - The last person takes 10 minutes to fill their bucket. Adding up all these times, the total time used is boxed{84} minutes.

answer:Definitions: For an n times n board filled with numbers (1, 2, 3, ldots, n^2), a square is considered "optimal" if the number in the square is greater than at least 2004 other numbers in both its row and its column. Step-by-Step Explanation: 1. **Row Greater Condition:** For a square to be "row-optimal", the number in the square must be greater than the numbers in at least 2004 other squares in the same row. Given that there are n squares in each row, this implies: [ text{Number of row-optimal squares per row} = n - 2004. ] 2. **Column Greater Condition:** Similarly, for an optimal square, the number must be greater than the numbers in at least 2004 other squares in the same column. Thus: [ text{Number of column-optimal squares per column} = n - 2004. ] 3. **Counting Optimal Squares:** Since an optimal square must fulfill both conditions simultaneously (row-optimal and column-optimal), the total number of optimal squares on the board will be: [ text{Total number of optimal squares} = n times (n - 2004). ] 4. **Construction of Optimal Board Configuration:** To ensure that a given configuration meets the criteria, consider filling the board with the numbers so that: - Each row has precisely 2004 numbers which are lower than the rest of the numbers in that row. - Each column also has precisely 2004 numbers which are lower than the rest of the numbers in that column. Specifically, label the square in the i-th row and j-th column with an asterisk (*) if: - j ranges from i to i+2003 (modulo n if necessary). 5. **Verification:** Each row and column will therefore contain exactly 2004 asterisks (*), marking those numbers that are not optimal. All other numbers do not have asterisks and fulfill the condition of being greater than 2004 other numbers in their respective rows and columns. 6. **Conclusion:** Therefore, the total number of optimal squares (those without asterisks) in such an n times n board configuration, where n geq 2004, is: [ boxed{n(n - 2004)} ]

answer:Let's use the principle of inclusion-exclusion to solve this problem. Let H be the number of students who play hockey, B be the number of students who play basketball, and N be the total number of students in the class. We are given: - N (total number of students) = 50 - H (students who play hockey) = 30 - B (students who play basketball) = 35 - Number of students who play neither = 10 We want to find the number of students who play both hockey and basketball, which we'll call HB. According to the principle of inclusion-exclusion, we can calculate HB as follows: HB = H + B - (H ∪ B) Where (H ∪ B) is the total number of students who play at least one of the sports (hockey or basketball). First, we need to find (H ∪ B), which is the total number of students who play at least one sport. We know that there are 10 students who play neither sport, so the number of students who play at least one sport is: (H ∪ B) = N - (Number of students who play neither) (H ∪ B) = 50 - 10 (H ∪ B) = 40 Now we can use the principle of inclusion-exclusion to find HB: HB = H + B - (H ∪ B) HB = 30 + 35 - 40 HB = 65 - 40 HB = 25 So, there are boxed{25} students who play both hockey and basketball.