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Letters: S, C, H, O, O, L (6 letters, O repeated 2 times). Permutations = \(\frac{6!}{2!}\) = \(\frac{720}{2}\) = 360.

The word HANDIER consists of 7 letters: H, A, N, D, I, E, R. All 7 letters are distinct (no repetitions). The number of distinct ways to arrange these 7 letters is the number of permutations of 7 different items, which is 7!. 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 Therefore, HANDIER can be arranged in 5040 ways.

The die has faces 1, 2, 3, 4, 5, 6. Even numbers are 2, 4, 6 (3 outcomes). Total outcomes are 6. Probability = \(\frac{3}{6}\) = \(\frac{1}{2}\).

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)

Area of triangle PQR = \(\frac{1}{2}\) x QR x Pc Where QR = base of triangle PQR, Pc = height of triangle PQR Area of triangle PQR = \(\frac{1}{2}\) x 20 x 13 = 10 x 13 = 130cm\(^2\).

Given that a * b = \(\frac{a + b}{\text{ab}}\) + (a - b) 3 * 2 = Given that a * b = \(\frac{3 + 2}{3.2}\) + (3 - 1) = \(\frac{5}{6}\) + 1 = \(\frac{5}{6}\) + \(\frac{6}{6}\) = \(\frac{11}{6}\)

Given: 125\(^{−\frac{1}{3}}\) \times 49\(^{−\frac{1}{2}}\) \times 10\(^0\) 125\(^{-1/3}\) = (5\(^3\))\(^{-1/3}\) = 5\(^{-1}\) = \(\frac{1}{5}\). 49\(^{-1/2}\) = (7\(^2\))\(^{-1/2}\) = 7\(^{-1}\) = \(\frac{1}{7}\). 10\(^0\) = 1. Product = \(\frac{1}{5}\) x \(\frac{1}{7}\) x 1 = \(\frac{1}{35}\).

Midpoint = [\(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\)] Midpoint = [\(\frac{(-3 + 5)}{2}\), \(\frac{(4 + 6)}{2}\)] = (\(\frac{2}{2}\), (\(\frac{10}{2}\)) = (1, 5).

Sin A = \(\sqrt{(1 - (\frac{12}{13})^2)}\) = \(\sqrt{(1 - \frac{144}{169}}\)) = \(\sqrt{(\frac{25}{169})}\) = \(\frac{5}{13}\). Tan A = (\(\frac{5}{13}\)) \(\div\) (\(\frac{12}{13}\)) = \(\frac{5}{12}\).

Mean = \(\frac{(0 + x + 2 + 3x + 6 + 4x + 8)}{4}\) = 4. Simplify numerator: 8x + 16. So \(\frac{(8x + 16)}{4}\) = 4. 8x + 16 = 16. 8x = 0. x = 0.

Convert the mixed number slope to an improper fraction: 3 \(\frac{1}{5} = \frac{16}{5}\) Use the point-slope form with the given point (3,15) y - 15 = \(\frac{16}{5}\)(x - 3) Multiply both sides by 5 to clear the fraction: 5(y−15) = 16(x−3) 5y − 75 = 16x − 48 Rearrange terms to standard form: 5y − 16x − 75 + 48 = 0 5y − 16x − 27 = 0

Given (\(\frac{3}{4} \div 2\frac{1}{4}\)) of 1 \(\frac{7}{11}\) (3 \(\frac{2}{3}\) - \(\frac{15}{6}\)) 1. Simplify the division in the first parentheses: \(\frac{3}{4} \div 2 \frac{1}{4} = \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} = \frac{3}{9} = \frac{1}{3}\). 2. Convert the mixed number: \(1 \frac{7}{11} = \frac{18}{11}\). 3. Simplify the subtraction in the second parentheses: \(3 \frac{2}{3} - \frac{15}{6} = \frac{11}{3} - \frac{5}{2} = \frac{22}{6} - \frac{15}{6} = \frac{7}{6}\). 4. The "of" indicates multiplication, so multiply the results: \(\frac{1}{3} \times \frac{18}{11} \times \frac{7}{6} = \frac{126}{198} = \frac{7}{11}\).