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Mean = \(\frac{ 13 + 16 + x + 18 + 21 + 2x + 35}{7}\) = 22 103 + 3x = 154 3x = 154 - 103 = 51 x = \(\frac{51}{3}\) = 17

To find the variance from the standard deviation: Given: Standard deviation = 12.34 Variance = (12.34)\(^2\) = 152.2756 Rounding to the nearest whole number: Variance ≈ 152. ​

mean = \(\frac{ 4 + 5 + 6 + 7 + 8}{5}\) = 6 x x - mean (x - mean)\(^2\) 4 4 - 6 = -2 4 5 5 - 6 = -1 1 6 6 - 6 = 0 0 7 7 - 6 = 1 1 8 8 - 6 = 2 4 Variance = \(\frac{10}{5}\) S.D = \(\sqrt{variance}\) = \(\sqrt{2}\)

Mode = 42 42 occurred thrice.

The measures of central tendency are: Mean, Median, and Mode while the measures of dispersion/spread are: Range, Variance, Standard Deviation, Interquartile Range (IQR), and Coefficient of Variation.

The number of ways is given by (n−1)! For six people = (6−1)! = 5! = 120 Thus, the number of ways to seat 6 people around the table is: 120

To form a committee of 2 girls and 3 boys from 5 girls and 7 boys: Step 1: Combinations for Girls = \(^5C_2\) = \(\frac{5!}{(5-2)!2!}\) = 10 Step 2: Calculate the Combinations for Boys = \(^7C_3\) = \(\frac{7!}{(7-3)!3!}\) = 35 the total number of ways to form the committee, multiply the combinations of girls and boys: 10 x 35 = 350ways The total number of ways to form the committee is: 350.

To form a committee of 5 from 7 males and 3 females with at least one female: Choose 1 female from 3: \(^3C_1\) = 3 Choose 4 males from 7: \(^7C_4\) = 35 Total ways = 3 \times 35 = 105 The total number of ways to select the committee is: 105ways.

Total Balls 7 + 4 = 11 Probability of Picking Two Balls of the Same Colour = RR or BB = \(\frac{7}{11} \times \frac{7}{11}\) + \(\frac{4}{11} \times \frac{4}{11}\) = \(\frac{49}{121}\) + \(\frac{16}{121}\) = \(\frac{65}{121}\)

Given the matrix \(P = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix},\) we want to find \( P^2 - 4P - I \), where \( I \) is the identity matrix: \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\) Calculate {\( P^2 \)} \(P^2 = P \times P = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}.\) Calculate {\( 4P \)} \(4P = 4 \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix}\) Calculate {\( P^2 - 4P \)} \(P^2 - 4P = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix} = \begin{bmatrix} 5 - 4 & 8 - 8 \\ 8 - 8 & 13 - 12 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) Calculate {\( P^2 - 4P - I \)} \(P^2 - 4P - I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) Thus, the result of \( P^2 - 4P - I \) is \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Given \(\tan \theta = \frac{8}{15}\) \(\sin \theta = \frac{8}{17}, \quad \cos \theta = \frac{15}{17}\) from pythagora's theorem. To find \(\frac{\sin \theta - \cos \theta}{\sin^2 \theta - \sin \theta}\) Numerator → \(\sin \theta - \cos \theta = \frac{8}{17} - \frac{15}{17} = \frac{-7}{17}\) Denominator → \(\sin^2 \theta = \left(\frac{8}{17}\right)^2 = \frac{64}{289}, \quad \sin^2 \theta - \sin \theta = \frac{64}{289} - \frac{136}{289} = \frac{-72}{289}\) \(\frac{\frac{-7}{17}}{\frac{-72}{289}} = \frac{7 \times 289}{17 \times 72} = \frac{119}{72}\)

t = \(\sqrt{(\frac{pq}{r} - r^2q)}\) Take the square of both sides t\(^2\) = \(\frac{pq}{r}\) - r\(^2\)q t\(^2\) = \(\frac{pq - r^3q}{r}\) cross multiply rt\(^2\) = pq - r\(^3\)q = q(p - r\(^3\)) q = \(\frac{rt^2}{p - r^3}\)

(\(\sqrt{a} +\sqrt{8a})^2\) = 54 + b\(\sqrt{2}\) (\(\sqrt{a} + \sqrt{8a}\))(\(\sqrt{a} + \sqrt{8a}\)) a + 4a\(\sqrt{2} + 8a = 54 + b\sqrt{2}\) 9a + 4a\(\sqrt{2}\) = 54 + b\(\sqrt{2}\) comparing both sides 9a = 54, so that a = 6 4a = b b = 4 x 6 = 24.

3.4 x 10\(^{-5}\) by 7.1 x 10\(^8\) = ( 3.4 x 7.1) x 10\(^{-5 + 8}\) = 24.14 x 10\(^3\) = 2.414 x 10\(^4\)

P = {p:1 < p < 20} Then, P = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} R = { r : 0 \(\leq\) 25, r is a multiple of 4 } Then R = { 0, 4, 8, 12, 16, 20, 24} Thus, P ∩ R = {4, 8, 12, 16}