This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the questions. Question 38: The diameter of a sphere is 12 cm. The radius r is half of the diameter. r = 12 cm2 = 6 cm The formula for the volume of a sphere is V = (4)/(3) r^3. We are given = (22)/(7). Step 1: Substitute the values into the volume formula. V = (4)/(3) × (22)/(7) × (6 cm)^3 V = (4)/(3) × (22)/(7) × 216 cm^3 Step 2: Perform the multiplication and division. V = (4 × 22 × 216)/(3 × 7) cm^3 V = (88 × 216)/(21) cm^3 V = (19008)/(21) cm^3 V ≈ 905.1428... cm^3 Step 3: Round the volume to the nearest cm^3. V ≈ 905 cm^3 The final answer is A. 905 cm^3. Question 39: A box contains 12 identical balls. Number of red balls = 5. Number of blue balls = 4. Number of green balls = Total balls - (Red balls + Blue balls) Green balls = 12 - (5 + 4) = 12 - 9 = 3 The probability of selecting a green ball is the number of green balls divided by the total number of balls. P(green) = Number of green ballsTotal number of balls = (3)/(12) Step 1: Simplify the probability. P(green) = (1)/(4) The final answer is D. (1)/(4). Question 40: Two balls are selected at random, one after the other, with replacement. Total balls = 12. Number of red balls = 5. Step 1: Calculate the probability of the first ball being red. P(1st red) = Number of red ballsTotal number of balls = (5)/(12) Step 2: Calculate the probability of the second ball being red. Since the first ball is replaced, the total number of balls and the number of red balls remain the same for the second draw. P(2nd red) = (5)/(12) Step 3: Calculate the probability that both balls are red. Since the events are independent (due to replacement), multiply their probabilities. P(both red) = P(1st red) × P(2nd red) P(both red) = (5)/(12) × (5)/(12) P(both red) = (25)/(144) The final answer is A. (25)/(144). Question 41: In the diagram, PQ is a straight line. The sum of angles on a straight line is 180^. From the diagram, the angles x, y, z, and m are on the straight line PQ. x + y + z + m = 180^ We are given the relationship m = (1)/(2)(x+y+z). Step 1: Express (x+y+z) in terms of m. From the given relationship, multiply both sides by 2: 2m = x+y+z Step 2: Substitute 2m for (x+y+z) into the equation for angles on a straight line. (2m) + m = 180^ 3m = 180^ Step 3: Solve for m. m = (180^)/(3) m = 60^ The final answer is B. 60^. Question 42: The points on a linear graph are given in the table. We need to find the gradient. The formula for the gradient G between two points (x_1, y_1) and (x_2, y_2) is: G = (y_2 - y_1)/(x_2 - x_1) Let's use the first two points from the table: (x_1, y_1) = (6.20, 3.90) (x_2, y_2) = (6.85, 5.20) Step 1: Substitute the coordinates into the gradient formula. G = (5.20 - 3.90)/(6.85 - 6.20) Step 2: Calculate the differences. G = (1.30)/(0.65) Step 3: Perform the division. G = 2 Let's verify with another pair of points, for example, the second and third points: (x_1, y_1) = (6.85, 5.20) (x_2, y_2) = (7.50, 6.50) G = (6.50 - 5.20)/(7.50 - 6.85) G = (1.30)/(0.65) G = 2 The gradient is consistent. The final answer is 2.