The diagram shows a flowerpot in the shape of a frustum of a cone. The diameters of the top and bottom ends are 64cm and 48cm respectively. If the slant height is 15cm, calculate the curved surface area of the pot.

Here are the solutions to the problems: Problem 7: Curved surface area of a frustum The diagram shows a flowerpot in the shape of a frustum of a cone. The diameters of the top and bottom ends are 64 cm and 48 cm respectively. If the slant height is 15 cm, calculate the curved surface area of the pot. Step 1: Identify the radii and slant height. Top diameter = 64 cm Top radius (R) = (64)/(2) = 32 cm. Bottom diameter = 48 cm Bottom radius (r) = (48)/(2) = 24 cm. Slant height (l) = 15 cm. Step 2: Apply the formula for the curved surface area of a frustum. The formula for the curved surface area of a frustum is A = (R + r) l. Using = (22)/(7) as in previous problems: A = (22)/(7) (32 + 24) (15) A = (22)/(7) (56) (15) Step 3: Calculate the area. A = 22 × ((56)/(7)) × 15 A = 22 × 8 × 15 A = 176 × 15 A = 2640 cm^2 The curved surface area of the pot is 2640 cm^2. --- Problem 8: Distribution of marks in an examination The table below shows the distribution of marks of candidates in an examination. a) Copy and complete the table. | Marks | Frequency | Marks less than | Cum. Freq. | | :------ | :-------- | :-------------- | :--------- | | 0-9 | 4 | 9.5 | 4 | | 10-19 | 7 | 19.5 | 11 | | 20-29 | 5 | 29.5 | 16 | | 30-39 | 10 | 39.5 | 26 | | 40-49 | 13 | 49.5 | 39 | | 50-59 | 20 | 59.5 | 59 | | 60-69 | 15 | 69.5 | 74 | | 70-79 | 13 | 79.5 | 87 | | 80-89 | 5 | 89.5 | 92 | | 90-99 | 1 | 99.5 | 93 | The completed table is shown above. The total frequency (N) is 93. b) Draw the cumulative frequency curve for distribution. To draw the cumulative frequency curve (ogive), plot the "Marks less than" values on the x-axis against the "Cum. Freq." values on the y-axis. The points to plot are: (0, 0) - (starting point for cumulative frequency) (9.5, 4) (19.5, 11) (29.5, 16) (39.5, 26) (49.5, 39) (59.5, 59) (69.5, 74) (79.5, 87) (89.5, 92) (99.5, 93) Connect these points with a smooth curve. c) Use your graph to estimate the median mark. Step 1: Determine the position of the median. The total number of candidates is N = 93. The median position is at (N)/(2) = (93)/(2) = 46.5^th candidate. Step 2: Estimate the median mark from the graph (using interpolation for precision). Locate 46.5 on the cumulative frequency axis. Draw a horizontal line to the curve, then a vertical line down to the marks axis. The 46.5th value falls within the 50-59 mark range. The cumulative frequency before this class (at 49.5 marks) is 39. The cumulative frequency at the end of this class (at 59.5 marks) is 59. The frequency of the 50-59 class is 20. The class width is 10. Using interpolation: Median = 49.5 + ((46.5 - 39)/(59 - 39)) × (59.5 - 49.5) Median = 49.5 + ((7.5)/(20)) × 10 Median = 49.5 + 0.375 × 10 Median = 49.5 + 3.75 Median = 53.25 The estimated median mark is 53.25. d) If the mark for distinction was 70%, what percentage of students passed with distinction? Step 1: Determine the number of students who scored 70 marks or more. A mark of 70% means a score of 70. From the cumulative frequency table, the number of students who scored less than 70 (i.e., up to 69.5 marks) is 74. ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._