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No of ways of choosing 1 man, 2 women = 5C1 x 3C2 No of ways of choosing 2 men, 1 woman = 5C2 x 3C1 Summing, => (5C1 x 3C2) + (5C2 x 3C1) = 15 + 30 = 45

In the given diagram, < PQS = \(\frac{1}{2}\) x 96º = 48º ( Therefore, < PST = 48º ( alternate segment).

Speed = 30 kmh\(^{-1}\) = \(\frac{30 \times 1000}{3600}\) = \(\frac{25}{3} ms^{-1}\) Time = \(\frac{Distance}{Speed}\) = \(\frac{200}{\frac{25}{3}}\) = \(\frac{200 \times 3}{25}\) = 24 seconds

The statement that is not true about a rectangle is: II and III only The diagonals of a rectangle do not intersect at right angles. and does not have four lines of symmetry but two.

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN OE = OG = r(radii) (r - 8)\(^2\) + 12\(^2\) = r\(^2\) r\(^2\) - 16r + 64 + 144 = r\(^2\) 16r = 64 + 144 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm

Area of trapezium = \(\frac{1}{2} (a + b)h\) \(120 = \frac{1}{2} (5 + 7) \times h\) \(120 = 6h\) \(h = 20 cm\)

The given figure is a kite. Here are the properties of a kite: Two Pairs of Adjacent Sides: A kite has two pairs of adjacent sides that are equal in length. Diagonals: The diagonals of a kite are perpendicular to each other. One diagonal bisects the other. Symmetry: A kite has one line of symmetry, which runs along the longer diagonal. Angles: The angles between the unequal sides are equal. Specifically, the angles opposite the equal sides are also equal. Therefore, ∠PQR = ∠PSR is correct.

\(\pi r^2\) = 38.5 \(\frac{22}{7} r^2\) = 38.5 r\(^2\) = \(\frac{38.5 \times 7}{22}\) r = \(\sqrt{12.25}\) = 3.5 diameter = 2r = 2 x 3.5 = 7cm

< T = < S = 50^\circ (OS = OT) < SOT = 180^\circ - 2(50^\circ) = 80^\circ < ROP = 80^\circ (vertically opposite angle) \(\therefore\) < OPQ = 180^\circ - 80^\circ = 100^\circ (adjacent angles)

Sum of interior angle in a hexagon = (6 - 2) x 180^\circ = 720^\circ \(\therefore\) 107^\circ + 2x^\circ + 150^\circ + 95^\circ + (2x - 15)^\circ + 123^\circ = 720^\circ 460 + 4x = 720 \(\implies\) 4x = 720 - 460 4x = 260^\circ \(\implies\) x = 65^\circ

circumference of the bicycle wheel \(= 2\pi r = 2 \times \frac{22}{7} \times 42 = 264cm = 2.64m\\\) Number of revolution \(=\frac{66}{2.64} = 25\)

Length of arc = \(\frac{\theta}{360} \times 2\pi r\) \(50 = \frac{75}{360} \times 2 \times \frac{22}{7} \times r\) \(r = \frac{50 \times 360 \times 7}{75 \times 2 \times 22}\) \(r = \frac{420}{11}\) = 38.18 cm \(\approxeq\) 38.2 cm (3 sig. figs)

In triangle ABC, \(\frac{BC}{AC} = \cos \theta\) \(\implies \cos \theta = \frac{5}{10} = 0.5\) \(\theta = \cos^{-1} (0.5) = 60^\circ\)

\(x^2 = 8^2 + 8^2\) \(x^2 = 64 + 64 = 128\) \(x = \sqrt{128}\) = \(8\sqrt{2}\) km

The bearing of Kofi from the building = 180^\circ + 45^\circ = 225^\circ