answer:1. **Understanding the Problem:** We have an 8 times 8 chessboard with alternating black and white squares. We need to place a 1 times 2 rectangular piece of paper on the board such that it overlaps the maximum number of black squares. 2. **Initial Configuration:** Consider the chessboard where each square is either black or white. The pattern alternates, so each row and column contains 4 black and 4 white squares. 3. **Placing the Rectangle:** The 1 times 2 rectangle can be placed either horizontally or vertically. We need to determine the placement that maximizes the overlap with black squares. 4. **Horizontal Placement:** - If the rectangle is placed horizontally, it covers two adjacent squares in the same row. - Since the chessboard alternates colors, the maximum number of black squares it can cover in this orientation is 1 (one black and one white). 5. **Vertical Placement:** - If the rectangle is placed vertically, it covers two adjacent squares in the same column. - Similar to the horizontal case, the maximum number of black squares it can cover in this orientation is also 1 (one black and one white). 6. **Optimal Placement:** - To maximize the number of black squares covered, we need to consider the placement of the rectangle such that it overlaps multiple rows and columns. - By placing the rectangle such that it spans across the intersection of four squares, we can cover more black squares. 7. **Covering Multiple Rows and Columns:** - Consider placing the rectangle such that its center is at the intersection of four squares. This way, the rectangle can overlap parts of up to 6 black squares. - This configuration is possible because the rectangle can be placed diagonally across the intersection of four squares, covering parts of each. 8. **Verification:** - To verify, consider the geometry of the placement. The rectangle's center is at the intersection of four squares, and it extends frac{1}{2} unit in one direction and 1 unit in the other direction. - This placement ensures that the rectangle overlaps parts of 6 black squares. 9. **Conclusion:** - It is not possible to overlap more than 6 black squares with a 1 times 2 rectangle on an 8 times 8 chessboard. - The maximum number of black squares that can overlap the paper is 6. The final answer is boxed{6}

answer:Emily bought 2 skirts at 15 each, so the total cost for the skirts is 2 * 15 = 30. She also bought art supplies for 20. Therefore, the total amount she spent at the store is the cost of the skirts plus the cost of the art supplies, which is 30 + 20 = boxed{50} .

answer:1. **Understanding the inequality**: The inequality x + 2y leq 6 defines a region in the first quadrant of the xy-plane, bounded by the line x + 2y = 6 and the coordinate axes. This region includes the triangle formed by the points (0,0), (6,0), and (0,3). 2. **Positive integer coordinates**: A point has positive integer coordinates if both x and y are positive integers. For each integer value of x, we determine the maximum possible integer y such that the inequality holds. 3. **Counting integer points**: We iterate over possible integer values for x and count valid y values: - For x = 1, 2y leq 4 implies y leq 2. Possible y values are 1, 2. - For x = 2, 2y leq 4 implies y leq 2. Possible y values are 1, 2. - For x = 3, 2y leq 3 implies y leq 1.5. Possible y value is 1. - For x = 4, 2y leq 2 implies y leq 1. Possible y value is 1. - For x = 5, 2y leq 1 implies y leq 0.5. No valid y values as we consider only positive integers. - For x = 6, 2y leq 0 implies y leq 0. No valid y values as we consider only positive integers. There are 6 points in total with valid positive integer coordinates: (1,1), (1,2), (2,1), (2,2), (3,1), (4,1). 4. **Conclusion**: The total number of points with positive integer coordinates satisfying the inequality x + 2y leq 6 is 6. The final answer is boxed{C}

answer:Let ( X ) represent the time until Chloe's choice, which has an exponential distribution with parameter ( lambda = 1/2 ). Similarly, let ( Y ) represent the time until Laurent’s choice, which follows an exponential distribution with parameter ( lambda = 1/4 ). The probability we need to find is ( P(Y > X) ). Given that ( X ) and ( Y ) are independent, we use the memoryless property of exponential distributions. Let ( P(Y > X) = int_0^infty P(Y > x)f_X(x) , dx ), where ( f_X(x) ) is the density function of ( X ). The density function of ( X ) is ( f_X(x) = frac{1}{2} e^{-x/2} ) for ( x geq 0 ) and that of ( Y ) is ( f_Y(y) = frac{1}{4} e^{-y/4} ) for ( y geq 0 ). Plugging these into the integral, we have: [ P(Y > X) = int_0^infty P(Y > x) cdot frac{1}{2} e^{-x/2} , dx = int_0^infty left( int_x^infty frac{1}{4} e^{-y/4} , dy right) cdot frac{1}{2} e^{-x/2} , dx. ] This simplifies to: [ P(Y > X) = int_0^infty e^{-x/4} cdot frac{1}{2} e^{-x/2} , dx = int_0^infty e^{-3x/4} cdot frac{1}{2} , dx. ] This integral evaluates to: [ = frac{1}{2} cdot frac{4}{3} = frac{2}{3}. ] Conclusion: The probability that Laurent's number is greater than Chloe's number is ( frac{2{3}} ). boxed{The correct answer is ( textbf{(C)} frac{2}{3} ).}