Table 2.2 shows the scores of candidates in an entrance examination into a college. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. (b) Use your curve to estimate the: Median mark, Upper and lower quartiles, Semi-inter quartile range.

Question

Asked on July 1, 2026Mathematics

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.

Accepted Answer · 2026-07-01T12:14:36.563Z

10. (a) Speed of X due to the rotation of the earth Step 1: Calculate the radius of the parallel of latitude (r). The latitude () is 38^. The Earth's radius (R) is 6400 km. r = R r = 6400 (38^) r = 6400 × 0.78801 r ≈ 5043.26 km Step 2: Calculate the angular speed () of the Earth. The Earth rotates 2 radians in 24 hours. = (2)/(24) rad/hour = ()/(12) rad/hour Step 3: Calculate the speed (v) of city X. v = r v = 5043.26 × (22/7)/(12) v = 5043.26 × (22)/(84) v ≈ 1321.0 km/h Step 4: Correct to 3 significant figures. v ≈ 1320 km/h The speed of X due to the rotation of the earth is 1320 km/h. 10. (b) Distance between X and Y along the great circle Step 1: Determine the angular difference in longitude (). Longitude of X = 50^ E. Longitude of Y = 130^ W. Since they are on opposite sides, add the longitudes. = 50^ + 130^ = 180^ Step 2: Convert the angular difference to radians. 180^ = radians Step 3: Calculate the distance (D) along the great circle. D = R _radians D = 6400 × Using = (22)/(7): D = 6400 × (22)/(7) D = (140800)/(7) D ≈ 20114.28 km Step 4: Correct to 3 significant figures. D ≈ 20100 km The distance between X and Y along the great circle is 20100 km. 10. (c) Distance between X and Y along the parallel of latitude Step 1: Identify the radius of the parallel of latitude (r) and the angular difference (). From part (a), r ≈ 5043.26 km. From part (b), = radians. Step 2: Calculate the distance (D_parallel) along the parallel of latitude. D_parallel = r _radians D_parallel = 5043.26 × Using = (22)/(7): D_parallel = 5043.26 × (22)/(7) D_parallel = (110951.72)/(7) D_parallel ≈ 15850.24 km Step 3: Correct to 3 significant figures. D_parallel ≈ 15900 km The distance between X and Y along the parallel of latitude is 15900 km. 11. (a) Copy and complete Table 2.1 for the relation y = 2x^2 - 5x - 6, for -2 x 5. Step 1: Calculate the missing y-values using the equation y = 2x^2 - 5x - 6. For x = -1: y = 2(-1)^2 - 5(-1) - 6 = 2(1) + 5 - 6 = 1. For x = 0: y = 2(0)^2 - 5(0) - 6 = 0 - 0 - 6 = -6. For x = 1: y = 2(1)^2 - 5(1) - 6 = 2 - 5 - 6 = -9. For x = 4: y = 2(4)^2 - 5(4) - 6 = 2(16) - 20 - 6 = 32 - 20 - 6 = 6. Step 2: Complete the table. | x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |---|----|----|---|---|---|---|---|---| | y | 12 | 1 | -6 | -9 | -8 | -3 | 6 | 19 | The completed table is shown above. 11. (b) Using a scale of 2 cm to represent 1 unit on x-axis and 2 cm to represent 5 units on y-axis, draw the graph of y = 2x^2 - 5x - 6 for -2 x 5. This step requires drawing a graph, which cannot be performed by an AI. You would plot the points from the table in part (a) on graph paper using the specified scale and draw a smooth curve through them. 11. (c) Use your graph to find the: (i) Roots of the equation 2x^2 - 5x - 6 = 0, correct to one decimal place. These are the x-intercepts of the graph y = 2x^2 - 5x - 6. You would read these values from your graph where the curve crosses the x-axis (y=0). Using the quadratic formula for 2x^2 - 5x - 6 = 0: x = -(-5) ± sqrt((-5)^2 - 4(2)(-6))2(2) x = 5 ± sqrt(25 + 48)4 x = 5 ± sqrt(73)4 x_1 = 5 + sqrt(73)4 ≈ (5 + 8.544)/(4) ≈ 3.386 ≈ 3.4 x_2 = 5 - sqrt(73)4 ≈ (5 - 8.544)/(4) ≈ -0.886 ≈ -0.9 The roots are approximately x = 3.4 and x = -0.9. (ii) Values of x for which 2x^2 - 5x - 8 = 0, correct to one decimal place. This equation can be rewritten as 2x^2 - 5x - 6 = 2. You would draw the line y = 2 on your graph and find the x-coordinates of the intersection points with the curve y = 2x^2 - 5x - 6. Using the quadratic formula for 2x^2 - 5x - 8 = 0: x = -(-5) ± sqrt((-5)^2 - 4(2)(-8))2(2) x = 5 ± sqrt(25 + 64)4 x = 5 ± sqrt(89)4 x_1 = 5 + sqrt(89)4 ≈ (5 + 9.434)/(4) ≈ 3.608 ≈ 3.6 x_2 = 5 - sqrt(89)4 ≈ (5 - 9.434)/(4) ≈ -1.108 ≈ -1.1 The values of x are approximately x = 3.6 and x = -1.1. (iii) Gradient of the tangent at the point where x = 2. You would draw a tangent line to the curve at x=2 on your graph and calculate its slope. Alternatively, using calculus: Step 1: Find the derivative of the function y = 2x^2 - 5x - 6. (dy)/(dx) = 4x - 5 Step 2: Substitute x = 2 into the derivative to find the gradient. Gradient = 4(2) - 5 = 8 - 5 = 3 The gradient of the tangent at the point where x=2 is 3. 12. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Step 1: Create the cumulative frequency table by adding the number of students for each class interval to the sum of the previous intervals. Use upper class boundaries for plotting the curve. | Marks (%) | No. of students (f) | Upper Class Boundary | Cumulative Frequency (cf) | |---|---|---|---| | 21-30 | 9 | 30.5 | 9 | | 31-40 | 20 | 40.5 | 29 | | 41-50 | 31 | 50.5 | 60 | | 51-60 | 42 | 60.5 | 102 | | 61-70 | 50 | 70.5 | 152 | | 71-80 | 32 | 80.5 | 184 | | 81-90 | 16 | 90.5 | 200 | The cumulative frequency table is shown above. Step 2: To draw the cumulative frequency curve (ogive), you would plot the cumulative frequency (y-axis) against the upper class boundaries (x-axis). The points to plot are (30.5, 9), (40.5, 29), (50.5, 60), (60.5, 102), (70.5, 152), (80.5, 184), (90.5, 200). Start the curve from (20.5, 0). 12. (b) Use your curve to estimate the: Total number of students (N) = 200. (i) Median mark The median is at the (N)/(2)-th position, which is (200)/(2) = 100-th student. Graphical estimation:* Locate 100 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line down to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 100th student falls in the 51-60 class. Median = L + ( (N)/(2) - cf_bf_m ) c Median = 50.5 + ( (100 - 60)/(42) ) 10 Median = 50.5 + ( (40)/(42) ) 10 ≈ 50.5 + 9.52 ≈ 60.02 The median mark is approximately 60.0. (ii) Upper and lower quartiles Lower Quartile (Q_1): Position = (N)/(4) = (200)/(4) = 50-th student. Graphical estimation:* Locate 50 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 50th student falls in the 41-50 class. Q_1 = L + ( (N)/(4) - cf_bf_Q1 ) c Q_1 = 40.5 + ( (50 - 29)/(31) ) 10 Q_1 = 40.5 + ( (21)/(31) ) 10 ≈ 40.5 + 6.77 ≈ 47.27 The lower quartile is approximately 47.3. Upper Quartile (Q_3): Position = (3N)/(4) = (3 × 200)/(4) = 150-th student. Graphical estimation:* Locate 150 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 150th student falls in the 61-70 class. Q_3 = L + ( (3N)/(4) - cf_bf_Q3 ) c Q_3 = 60.5 + ( (150 - 102)/(50) ) 10 Q_3 = 60.5 + ( (48)/(50) ) 10 ≈ 60.5 + 9.6 ≈ 70.1 The upper quartile is approximately 70.1. (iii) Semi-inter quartile range (SIQR) SIQR = (Q_3 - Q_1)/(2) SIQR = (70.1 - 47.3)/(2) SIQR = (22.8)/(2) SIQR = 11.4 The semi-inter quartile range is 11.4. Got more? Send 'em!

Accepted Answer · 2026-07-01T12:14:36.563Z

10. (a) Speed of X due to the rotation of the earth Step 1: Calculate the radius of the parallel of latitude (r). The latitude () is 38^. The Earth's radius (R) is 6400 km. r = R r = 6400 (38^) r = 6400 × 0.78801 r ≈ 5043.26 km Step 2: Calculate the angular speed () of the Earth. The Earth rotates 2 radians in 24 hours. = (2)/(24) rad/hour = ()/(12) rad/hour Step 3: Calculate the speed (v) of city X. v = r v = 5043.26 × (22/7)/(12) v = 5043.26 × (22)/(84) v ≈ 1321.0 km/h Step 4: Correct to 3 significant figures. v ≈ 1320 km/h The speed of X due to the rotation of the earth is 1320 km/h. 10. (b) Distance between X and Y along the great circle Step 1: Determine the angular difference in longitude (). Longitude of X = 50^ E. Longitude of Y = 130^ W. Since they are on opposite sides, add the longitudes. = 50^ + 130^ = 180^ Step 2: Convert the angular difference to radians. 180^ = radians Step 3: Calculate the distance (D) along the great circle. D = R _radians D = 6400 × Using = (22)/(7): D = 6400 × (22)/(7) D = (140800)/(7) D ≈ 20114.28 km Step 4: Correct to 3 significant figures. D ≈ 20100 km The distance between X and Y along the great circle is 20100 km. 10. (c) Distance between X and Y along the parallel of latitude Step 1: Identify the radius of the parallel of latitude (r) and the angular difference (). From part (a), r ≈ 5043.26 km. From part (b), = radians. Step 2: Calculate the distance (D_parallel) along the parallel of latitude. D_parallel = r _radians D_parallel = 5043.26 × Using = (22)/(7): D_parallel = 5043.26 × (22)/(7) D_parallel = (110951.72)/(7) D_parallel ≈ 15850.24 km Step 3: Correct to 3 significant figures. D_parallel ≈ 15900 km The distance between X and Y along the parallel of latitude is 15900 km. 11. (a) Copy and complete Table 2.1 for the relation y = 2x^2 - 5x - 6, for -2 x 5. Step 1: Calculate the missing y-values using the equation y = 2x^2 - 5x - 6. For x = -1: y = 2(-1)^2 - 5(-1) - 6 = 2(1) + 5 - 6 = 1. For x = 0: y = 2(0)^2 - 5(0) - 6 = 0 - 0 - 6 = -6. For x = 1: y = 2(1)^2 - 5(1) - 6 = 2 - 5 - 6 = -9. For x = 4: y = 2(4)^2 - 5(4) - 6 = 2(16) - 20 - 6 = 32 - 20 - 6 = 6. Step 2: Complete the table. | x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |---|----|----|---|---|---|---|---|---| | y | 12 | 1 | -6 | -9 | -8 | -3 | 6 | 19 | The completed table is shown above. 11. (b) Using a scale of 2 cm to represent 1 unit on x-axis and 2 cm to represent 5 units on y-axis, draw the graph of y = 2x^2 - 5x - 6 for -2 x 5. This step requires drawing a graph, which cannot be performed by an AI. You would plot the points from the table in part (a) on graph paper using the specified scale and draw a smooth curve through them. 11. (c) Use your graph to find the: (i) Roots of the equation 2x^2 - 5x - 6 = 0, correct to one decimal place. These are the x-intercepts of the graph y = 2x^2 - 5x - 6. You would read these values from your graph where the curve crosses the x-axis (y=0). Using the quadratic formula for 2x^2 - 5x - 6 = 0: x = -(-5) ± sqrt((-5)^2 - 4(2)(-6))2(2) x = 5 ± sqrt(25 + 48)4 x = 5 ± sqrt(73)4 x_1 = 5 + sqrt(73)4 ≈ (5 + 8.544)/(4) ≈ 3.386 ≈ 3.4 x_2 = 5 - sqrt(73)4 ≈ (5 - 8.544)/(4) ≈ -0.886 ≈ -0.9 The roots are approximately x = 3.4 and x = -0.9. (ii) Values of x for which 2x^2 - 5x - 8 = 0, correct to one decimal place. This equation can be rewritten as 2x^2 - 5x - 6 = 2. You would draw the line y = 2 on your graph and find the x-coordinates of the intersection points with the curve y = 2x^2 - 5x - 6. Using the quadratic formula for 2x^2 - 5x - 8 = 0: x = -(-5) ± sqrt((-5)^2 - 4(2)(-8))2(2) x = 5 ± sqrt(25 + 64)4 x = 5 ± sqrt(89)4 x_1 = 5 + sqrt(89)4 ≈ (5 + 9.434)/(4) ≈ 3.608 ≈ 3.6 x_2 = 5 - sqrt(89)4 ≈ (5 - 9.434)/(4) ≈ -1.108 ≈ -1.1 The values of x are approximately x = 3.6 and x = -1.1. (iii) Gradient of the tangent at the point where x = 2. You would draw a tangent line to the curve at x=2 on your graph and calculate its slope. Alternatively, using calculus: Step 1: Find the derivative of the function y = 2x^2 - 5x - 6. (dy)/(dx) = 4x - 5 Step 2: Substitute x = 2 into the derivative to find the gradient. Gradient = 4(2) - 5 = 8 - 5 = 3 The gradient of the tangent at the point where x=2 is 3. 12. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Step 1: Create the cumulative frequency table by adding the number of students for each class interval to the sum of the previous intervals. Use upper class boundaries for plotting the curve. | Marks (%) | No. of students (f) | Upper Class Boundary | Cumulative Frequency (cf) | |---|---|---|---| | 21-30 | 9 | 30.5 | 9 | | 31-40 | 20 | 40.5 | 29 | | 41-50 | 31 | 50.5 | 60 | | 51-60 | 42 | 60.5 | 102 | | 61-70 | 50 | 70.5 | 152 | | 71-80 | 32 | 80.5 | 184 | | 81-90 | 16 | 90.5 | 200 | The cumulative frequency table is shown above. Step 2: To draw the cumulative frequency curve (ogive), you would plot the cumulative frequency (y-axis) against the upper class boundaries (x-axis). The points to plot are (30.5, 9), (40.5, 29), (50.5, 60), (60.5, 102), (70.5, 152), (80.5, 184), (90.5, 200). Start the curve from (20.5, 0). 12. (b) Use your curve to estimate the: Total number of students (N) = 200. (i) Median mark The median is at the (N)/(2)-th position, which is (200)/(2) = 100-th student. Graphical estimation:* Locate 100 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line down to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 100th student falls in the 51-60 class. Median = L + ( (N)/(2) - cf_bf_m ) c Median = 50.5 + ( (100 - 60)/(42) ) 10 Median = 50.5 + ( (40)/(42) ) 10 ≈ 50.5 + 9.52 ≈ 60.02 The median mark is approximately 60.0. (ii) Upper and lower quartiles Lower Quartile (Q_1): Position = (N)/(4) = (200)/(4) = 50-th student. Graphical estimation:* Locate 50 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 50th student falls in the 41-50 class. Q_1 = L + ( (N)/(4) - cf_bf_Q1 ) c Q_1 = 40.5 + ( (50 - 29)/(31) ) 10 Q_1 = 40.5 + ( (21)/(31) ) 10 ≈ 40.5 + 6.77 ≈ 47.27 The lower quartile is approximately 47.3. Upper Quartile (Q_3): Position = (3N)/(4) = (3 × 200)/(4) = 150-th student. Graphical estimation:* Locate 150 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value. Analytical estimation (linear interpolation):* The 150th student falls in the 61-70 class. Q_3 = L + ( (3N)/(4) - cf_bf_Q3 ) c Q_3 = 60.5 + ( (150 - 102)/(50) ) 10 Q_3 = 60.5 + ( (48)/(50) ) 10 ≈ 60.5 + 9.6 ≈ 70.1 The upper quartile is approximately 70.1. (iii) Semi-inter quartile range (SIQR) SIQR = (Q_3 - Q_1)/(2) SIQR = (70.1 - 47.3)/(2) SIQR = (22.8)/(2) SIQR = 11.4 The semi-inter quartile range is 11.4. Got more? Send 'em!

Table 2.2 shows the scores of candidates in an entrance examination into a college. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. (b) Use your curve to estimate the: Median mark, Upper and lower quartiles, Semi-inter quartile range.

Need help with your own homework?

More Mathematics Questions

Table 2.2 shows the scores of candidates in an entrance examination into a college. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. (b) Use your curve to estimate the: Median mark, Upper and lower quartiles, Semi-inter quartile range.

Need help with your own homework?

More Mathematics Questions