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midpoint(x) = \(\frac{x_1 + x_2}{2}\) midpoint(y) = \(\frac{y_1 + y_2}{2}\) Given (9, 6) midpoint(x) = \(\frac{x + 10}{2}\) = 9 x + 10 = 2 x 9 = 18 x = 18 - 10 = 8.

From the diagram above: Tan 30º = \(\frac{15}{x}\) \(\frac{\sqrt{3}}{3}\) = \(\frac{15}{x}\) x\(\sqrt{3}\) = 3 x 15 = 45 x = \(\frac{45\sqrt{3}}{3}\) = 15\(\sqrt{3}\)m ( this is after rationalizing the denominator)

\(\frac{\text{dy}}{dx}\) = (5x + 1)\(^4\) = 20(5x + 1)\(^3\) let u = (5x + 1) = \(\frac{du}{dx}\) = 5 y = u\(^4\) then \(\frac{dy}{du}\) = 4u\(^3\) Therefore, \(\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}\) = 5 x 4u\(^3\) = 20(5x + 1)\(^3\)

Cos25º - Sin 25º The differentiation of a constant = 0, But based on the Options provided Cos 25º when differentiated = -Sin 25º and Sin25º = + Cos25º Putting these together, we have - Sin25º - Cos25º = - ( Sin25º + Cos25º)

MODE = 3, the most frequent number.

Total ages of four female teachers = 4 x 40 = 160 Total ages of eight male teachers = 8 x 25 = 200 Total ages of all the teachers in the school = 160 + 200 = 360 Average ages of all the teachers = \(\frac{\text{ total ages of all teachers}}{\text{total number of teachers}}\) = \(\frac{360}{12}\) = 30.

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 = \(\frac{7}{11} \times \frac{7}{11}\) + \(\frac{4}{11} \times \frac{4}{11}\) = \(\frac{49}{121}\) + \(\frac{16}{121}\) = \(\frac{65}{121}\)