What is the class width of the distribution? Calculate the mean marks of the distribution.

Here are the solutions to the problems: Part b) Class width Step 1: Identify two consecutive class intervals. The given class intervals are 41-49 and 50-58. Step 2: Calculate the class width. The class width can be found by subtracting the lower limit of one class from the lower limit of the next class, or by using the formula (upper limit - lower limit) + 1 for a single class. Using the lower limits: 50 - 41 = 9. Using the first class: (49 - 41) + 1 = 8 + 1 = 9. The class width is 9. Part c) Calculate the mean marks of the distribution. To calculate the mean of a frequency distribution, we need the class mark (x) and frequency (f) for all classes in the distribution. The formula for the mean is x = ( fx)/( f). From the provided image, only the data for the first class interval (41-49) is visible: Class Mark (x) = 45 Frequency (f) = 10 fx = 450 The frequencies for the other class intervals (e.g., 50-58, and 86-94) are missing, as are the total sums f and fx. Therefore, it is not possible to calculate the mean marks of the entire distribution with the information provided. Question 19 A) Inequalities Step 1: Identify the boundary lines for the inequalities. The given inequalities are x > 0, y > 0, and x + 2y 6. The boundary lines are: • x = 0 (the y-axis) • y = 0 (the x-axis) • x + 2y = 6 Step 2: Find two points for the line x + 2y = 6. • If x = 0, then 2y = 6 y = 3. Point: (0, 3). • If y = 0, then x = 6. Point: (6, 0). Step 3: Describe the region. The region satisfying x > 0 is to the right of the y-axis. The region satisfying y > 0 is above the x-axis. The region satisfying x + 2y 6 is below or on the line x + 2y = 6. Combining these, the region is a triangle in the first quadrant, bounded by the x-axis, the y-axis, and the line x + 2y = 6. The vertices of this triangular region are (0,0), (6,0), and (0,3). The region includes the line x+2y=6 but not the axes themselves (due to strict inequalities x>0, y>0). Step 4: State the integral values of x and y that satisfy the inequalities. We need integer values for x and y such that x 1, y 1, and x + 2y 6. • If y = 1: x + 2(1) 6 x + 2 6 x 4. Since x 1, the possible integer values for x are 1, 2, 3, 4. Points: (1,1), (2,1), (3,1), (4,1). • If y = 2: x + 2(2) 6 x + 4 6 x 2. Since x 1, the possible integer values for x are 1, 2. Points: (1,2), (2,2). • If y = 3: x + 2(3) 6 x + 6 6 x 0. Since x 1, there are no integer values for x when y=3. The integral values of x and y that satisfy the inequalities are: (1,1), (2,1), (3,1), (4,1), (1,2), (2,2) Question 19 B) Rectangle perimeter and diagonal Step 1: Set up equations based on the given information. Let the length of the rectangle be L and the width be W. The perimeter P = 2(L + W) = 46 cm. The diagonal D = 17 cm. By the Pythagorean theorem, L^2 + W^2 = D^2. Step 2: Formulate a system of equations. From the perimeter: 2(L + W) = 46 L + W = 23 (1) From the diagonal: L^2 + W^2 = 17^2 L^2 + W^2 = 289 (2) Step 3: Solve the system of equations. From equation (1), express L in terms of W: L = 23 - W Substitute this into equation (2): (23 - W)^2 + W^2 = 289 529 - 46W + W^2 + W^2 = 289 2W^2 - 46W + 529 - 289 = 0 2W^2 - 46W + 240 = 0 Divide the entire equation by 2: W^2 - 23W + 120 = 0 Step 4: Solve the quadratic equation for W. We can factor the quadratic equation. We need two numbers that multiply to 120 and add up to -23. These numbers are -8 and -15. (W - 8)(W - 15) = 0 This gives two possible values for W: W = 8 or W = 15 Step 5: Determine the corresponding length for each width. • If W = 8 cm, then L = 23 - 8 = 15 cm. • If W = 15 cm, then L = 23 - 15 = 8 cm. Conventionally, the width is considered the shorter side of the rectangle. Therefore, the width of the rectangle is 8 cm. Drop the next question! 📸