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An ogive is constructed using the cumulative frequencies from a frequency distribution table. Collect Data: Organize your data into a frequency distribution table. Calculate Cumulative Frequency: Add frequencies cumulatively for each class interval. Determine Class Boundaries: Identify the upper boundaries of the class intervals. Plot Points: Plot cumulative frequencies against the upper class boundaries. Draw the Ogive: Connect the points with a smooth curve, starting from (0, 0). Label the Axes: Clearly label the x-axis (upper boundaries) and y-axis (cumulative frequency).

A = P + 1 I = A - P = 10,400 - P Now using I = \(\frac{P \times T \times R}{100}\) i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\) = 100(10,400 - P) = 30P 10(10,400 - P) = 3P 104,000 - 10P = 3P 104,000 - 10P = 3P 104,000 = 3P + 10P = 104,000 = 13P P = \(\frac{104,000}{100}\) P = D8,000

Let: (x1, y1) = (1, 2) (x2, y2) = (5, 8) The gradient m of \(\bar{PQ}\) is given by m = \(\frac{y_2 y_1}{x_2 - x_1}\) = \(\frac{8 - 2}{5 - 1}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)

sin 60o = x + 0.5 0o(given) 0.8660 = x + 0.5 0.8660 - 0.5 = x x = 0.3660 cos\(\theta\) = x(given) cos\(\theta\) = 0.3660 Hence, \(\theta\) = cos-1(0.3660) = 68.53o = 69o (nearest degree)

In In the diagram given, \(\alpha\) = 22º (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles) < PSQ + 113º + \(\beta\) = 180º (sum of angles of a \(\Delta\)) < PSQ + 113º + 22º = 180º < PSQ = 180º - (113 + 22)º 45º

In the diagram above, \(\alpha\) = 2mo (angle at centre = 2 x angle at circumference) \(\alpha\) + 10mo = 360o (angle at circumference) \(\alpha\) + 10mo = 360o(angles round a point) 2mo + 10mo = 360o 12mo = 360o mo = \(\frac{360^o}{12}\) = 30o

By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\) \(\frac{8}{3} = \frac{20}{h}\) h = \(\frac{3 \times 20}{8}\) = 7.5m

In the diagram, < TUQ + 60o(corresp. angles) < QTU = 48o (alternate angles) < QU + 60o + 48o = 180o(sum of angles of a \(\Delta\)) < TQU = 180o - 108o = 72o

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\) \(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\)) = 30o In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1 \(\beta\) = tan-1(1) = 45o Hence, o + 45o = 75o

sectoral angle representing food = 360o - (80 + 70 + 90)o = 120o Amount spent on food = \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000 = Le 3,600,000

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2 Area of \(\Delta\) QMN = Area of \(\Delta\) QNP = Area of \(\Delta\) PNO (triangles between the same parallels) Hence, area of the trapezium 3 x area of \(\Delta\) QNP = 3 x 27 = 81cm2

Perimeter of a sector = 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3 64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3 22 = \(\frac{33\theta}{90}\) \(\theta = \frac{22 \times 30}{11}\) = 60o

In the diagram, o(angles in same segment) o - o o (same reason as above) o

From the diagram sin 70o = \(\frac{x}{10}\) x = 10sin 70o = 9.3969 Hence, length of chord MN = 2x = 2 x 9.3969 = 18.7938 = 19cm (nearest cm)

In the diagram, \(\beta\) = p(vertically opposite angles) m + \(\beta\) = n(sum of interior opp. angles) m + p = n