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1: First survives, second dies: \( p \cdot q \) 2: First dies, second survives: \( q \cdot p \) The total probability can be expressed as: \(\text{Total Probability} = pq + qp = 2pq\) Since \( 2pq \) is not an option, the closest form is \( pq \).

Sum \ge 10 ⇒ 6 favourable outcomes out of 36 Pr(sum at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Total balls = 2 + 6 + 5 = 13 Pr(black) = \(\frac{5}{13}\)

a. Total number of cards = 52 There are 4 fives and 2 red nines A card is drawn at random Pr(Either a five or a red nine) = Pr(five) + Pr(red nine) = \(\frac{4}{52}\) + \(\frac{2}{52}\) = \(\frac{6}{52}\) = \(\frac{3}{26}\) b. see diagram above ∠Q = 90^\circ \Delta PQR is a right-angled triangle. PR\(^2\) = PQ\(^2\) + QR\(^2\) PR\(^2\) = 5\(^2\) + 3\(^2\) = 25 + 9 PR\(^2\) = 34 = > PR = \(\sqrt{34}\) = 5.831 Let angle P = \theta Therefore, \(\frac{5.831}{sin 90^\circ}\) = \(\frac{3}{sin \theta}\) 5.831 sin \theta = 3 Therefore, Sin \theta = \(\frac{3}{5.831}\) = 0.5145 \(\theta\) = Sin\(^{-1}\)(0.5145) = 30.96º Bearing of R from P = 150^\circ - \theta = 150^\circ - 30.96^\circ = 119^\circ (to the nearest degree)

\(\frac{16^{2 + x}}{4} = 64^x\) 16\(^{2 + x} = 4 \times 64^x\) 4\(^{2(2 + x)} = 4^1 \times 4^{3(x)}\) (Equate coefficients since the bases are the same) 2(2 + x) = 1 + 3x 4 + 2x = 1 + 3x 2x - 3x = 1 - 4 -x = -3 x = 3.

Length of Arc = \(\frac{\theta}{360}\) x 2\(\pi\)r Length of arc = 33cm, diameter = 28cm → r = 14cm 33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7}\times14\) \(\theta\) = \(\frac{33 \times 180}{22 \times 2}\) \(\theta\) = 135º

Length of a chord = 2rSin\(\frac{\theta}{2}\) \(\theta\) = 60º not 30º and radius = 9cm Therefore, Length of the chord = 2 x 9 Sin\(\frac{60º}{2}\) = 2 x 9 x Sin 30º = 2 x 9 x \(\frac{1}{2}\) = 9cm.

The area of a triangle given two sides and the included angle = \(\frac{1}{2}\) x a x b x Sin\(\theta\) a = b = 8cm \(\theta\) = 60º Area = \(\frac{1}{2}\) x 8 x 8 x Sin 60º = \(\frac{1}{2}\) x 64 x \(\frac{\sqrt{3}}{2}\) ( Sin 60º = \(\frac{\sqrt{3}}{2}\) ) = \(\frac{64\sqrt{3}}{4}\) = 16\(\sqrt{3}cm^2\)

The construction of angle 30º (If you take a 60º angle and bisect it, you would divide it into two angles of 30º each.)

\(\overline{MN} || \overline{KL}\), \(\triangle\) LXK is similar to \(\triangle\)MXN So, the ratio of the areas of the two similar triangles equals the square of the ratio of their corresponding sides Therefore, \(\frac{ Area of \triangle MXN}{Area of \triangle LXK} = \frac{12^2}{9^2}\) \(\frac{16}{Area of \triangle LXK} = \frac{144}{81}\) Area of \(\triangle\)LXK = \(\frac{ 16 \times 81}{144} = 9cm^2\)

10111.11\(_{two}\) convert the number to base 10. 10111.11\(_{two}\) = 1 x 2\(^4\) + 0 x 2\(^3\) + 1 x \(^2\) + 1 x 2\(^1\) + 1 x 2\(^0\) + 1 x 2\(^{-1}\) + 1 x 2\(^{-2}\) = 16 + 0 + 4 + 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) = 23 + \(\frac{3}{4}\) = 23\(\frac{3}{4}\)

Subtract 14256\(_{seven}\) from 20045\(_{seven}\) = 2456\(_{seven}\) 2 0 0 4 5 1 4 2 5 6 0 2 4 5 6

16\(^x\) = 0.25 16\(^x\) = \(\frac{25}{100} = \frac{1}{4}\) 4(\(^2)^x\) = 4\(^{-1}\) 2x = -1 x = \(\frac{-1}{2}\)

\(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) multiply the denominator and the numerator by the conjugate of 3 + \(\sqrt{7}\) → 3 - \(\sqrt{7}\) \(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) x \(\frac{3 - \sqrt{7}}{3 - \sqrt{7}}\) \(\frac{15 - 5\sqrt{7} + 3\sqrt{7} - 7}{3^2 - (\sqrt{7})^2}\) \(\frac{8 - 2\sqrt{7}}{9 - 7}\) = \(\frac{ 8 - 2\sqrt{7}}{2}\) = 4 - \(\sqrt{7}\)

65 = 30 -x + x + 45 - x 65 - 75 = 10 = x ( number of students offering both subjects) Pr( \(\text{students offering both subjects}\)) = \(\frac{10}{65}\) = \(\frac{2}{13}\)