answer:1. **Calculate the True Average of the Original Scores**: Let the 50 scores be x_1, x_2, dots, x_{50}. The true average is: [ text{True Average} = frac{x_1 + x_2 + cdots + x_{50}}{50} ] Denote this average as A. Thus, ( A = frac{x_1 + x_2 + cdots + x_{50}}{50} ). 2. **Calculate the New Average Including the Additional Score**: Including an additional score of 100 results in 51 scores: [ x_1 + x_2 + cdots + x_{50} + 100 ] The sum of these 51 scores is: [ x_1 + x_2 + cdots + x_{50} + 100 = x_1 + x_2 + cdots + x_{50} + 100 ] The new average, A', is: [ A' = frac{x_1 + x_2 + cdots + x_{50} + 100}{51} ] Substituting for the sum of the original scores: [ A' = frac{50A + 100}{51} ] 3. **Calculate the Ratio of the New Average to the True Average**: [ frac{A'}{A} = frac{frac{50A + 100}{51}}{A} = frac{50A + 100}{51A} = frac{50 + frac{100}{A}}{51} ] # Conclusion: The ratio of the new average to the original average depends on the value of A and is given by frac{50 + frac{100{A}}{51}}. The final answer is boxed{C}

answer:In the coordinate plane, let A = (0,2), B = (2,-2), and P = (x,x). Then [AP = sqrt{x^2 + (x-2)^2}] and [BP = sqrt{(x-2)^2 + (x+2)^2},] so we want to minimize AP + BP, with P lying on the line y = x. By expanding the expressions, we get: [AP = sqrt{x^2 + (x-2)^2} = sqrt{x^2 + x^2 - 4x + 4} = sqrt{2x^2 - 4x + 4}] [BP = sqrt{(x-2)^2 + (x+2)^2} = sqrt{x^2 - 4x + 4 + x^2 + 4x + 4} = sqrt{2x^2 + 8}] Now, we minimize AP + BP: [AP + BP = sqrt{2x^2 - 4x + 4} + sqrt{2x^2 + 8}] By the Triangle Inequality, [AP + BP ge AB = sqrt{(2-0)^2 + (-2-2)^2} = sqrt{4 + 16} = sqrt{20} = 2sqrt{5}.] Equality occurs when P is the intersection of the line y = x and the line through A and B. Solving the equations of the lines: [ y = x ] [ (y-2) = -(x-2) Rightarrow y = -x + 4 ] Setting y = x, we get: [ x = -x + 4 Rightarrow 2x = 4 Rightarrow x = 2 ] Thus, P = (2,2), and checking shows this point satisfies the condition for equality in the triangle inequality. Conclusion with boxed answer: [ boxed{2sqrt{5}} ]

answer:1. **Identify Satisfactory Grades**: Grades A, B, and C are defined as satisfactory. 2. **Count the Number of Satisfactory Grades**: - Number of students with grade A = 6 - Number of students with grade B = 5 - Number of students with grade C = 4 - Total number of satisfactory grades = (6 + 5 + 4 = 15). 3. **Determine the Total Number of Students**: - Number of students with unsatisfactory grades (grades D and F) = 2 (D) + 6 (F) = 8 - Total number of students = Number of satisfactory grades + Number of unsatisfactory grades = (15 + 8 = 23). 4. **Calculate the Fraction of Satisfactory Grades**: - Fraction of satisfactory grades = (frac{text{Number of satisfactory grades}}{text{Total number of students}} = frac{15}{23}). 5. **Conclusion** and Simplify: - The fraction cannot be simplified further as 15 and 23 have no common factors besides 1. - Therefore, the final answer is (frac{15{23}}). The final answer is boxed{B) (frac{15}{23})}

answer:1. Suppose the first checker A returns to its initial position after having visited every square exactly once. This means that before A returned to its initial position, it must have moved away from it and subsequently visited each of the remaining squares. 2. Consider the moves of checker A. It must have traversed all squares on the board, which means that at some point, A must have left its initial position, hence causing some other checkers to also move if they were initially occupying positions into the path of A. 3. When A started moving, each of the positions on the board needed to be vacated at least once to allow A to complete its journey across all squares. Therefore, other checkers must have also moved to different positions on the board during A's journey. 4. Let's visualize the situation before A returns to its starting position. For A to have made its complete tour of the board and returned to its initial position, there must have been an intermediate point in time where A was away from its starting position, and all other checkers also must have left their starting positions since the total number of checkers and the number of free spaces are fixed. 5. Specifically, the instance just before A returns to its starting position is critical. At this moment, A is about to step back to its initial position, but until A does so, it stands on a different square. Similarly, since all checkers are to return to their starting points only after visiting all squares, they cannot have manifested their initial positions in this instance. This means that just before any checker returns to its starting position after covering all squares, no checker, including A, can be standing on its initial position. 6. Consequently, it implies there was always at least one moment when none of the checkers were standing on their initial positions. Because if there was a checker on its initial position while A is not, it would negate the movement dynamic described, which affects the overall traversal of the remaining checkers. Conclusion: Therefore, by logical implication and traversal constraints of chessboard positions, there must be a moment in which none of the checkers were standing on their starting positions. blacksquare