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Mean = \(\frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9}\) = \(\frac{(0 + 2 + 4 + 3 + 36)}{15}\) = \(\frac{45}{15}\) =3

Let the 4th angle be = x ∴ x + 45 + 90 + 135 = 360 x + 270 = 360 x = 360 - 270 x = 90 ∴ smallest angle = 45o 45o = N28.00 1o = 28.00/45 135o = (28.00/45) * (135/1) = N84.00

To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we start by using the formulas for permutations and combinations: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) \(^{n}P_{3} = \frac{n!}{(n-3)!}\) \(^{n}C_{4} = \frac{n!}{4!(n-4)!}\) \(\frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0\) Factor out \(n!\) Factoring \(n!\) from both terms gives: \(n! \left(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!}\right) = 0\) Since \(n! \neq 0\), we can simplify to: \(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!} = 0\) Recall that \(4! = 24\), so: \(\frac{6}{4!(n-4)!} = \frac{6}{24(n-4)!} = \frac{1}{4(n-4)!}\) The equation now is: \(\frac{1}{(n-3)!} - \frac{1}{4(n-4)!} = 0\) Multiply through by \(4(n-3)!(n-4)!\) gives: \(4(n-4)! - (n-3)! = 0\) Recall that \((n-3)! = (n-3)(n-4)!\), so: \(4(n-4)! - (n-3)(n-4)! = 0\) \((n-4)!\) Factoring out \((n-4)!\) gives: \((n-4)! \left(4 - (n-3)\right) = 0\) Since \((n-4)! \neq 0\), we have: \(4 - (n-3) = 0\) Solving for \(n\): \(n - 3 = 4 \implies n = 7\) The value of n is: 7

\(^{4}C_2 \times ^{3}C_1\) = \(\left(\frac{4!}{(4-2)!2!}\right)\times\left(\frac{3!}{(3-1)!1!}\right)\) = \(\frac{4!}{2!2!}\times \frac{3!}{2!}\) = \(\frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{3 \times 2!}{2!} = 18\)

Black balls = 5 Red balls = 3 P(B) = \(\frac{5}{8}\), P(R) = \(\frac{3}{8}\) P( a Black and a Red) = black ball and red ball or red ball and black ball i.e BR + RB = (\(\frac{5}{8}\)) * (\(\frac{3}{7}\)) + (\(\frac{3}{8}\)) * (\(\frac{5}{7}\)) = (\(\frac{15}{56}\)) + (\(\frac{15}{56}\)) = \(\frac{30}{56}\) = \(\frac{15}{28}\)

12 + 20 + x + 21 + x-1 +28 = 120 2x + 80 = 120 2x = 120 - 80 2x = 40 x = 20

Since PQ = QR, \Delta PQR is a right angle isosceles triangle ∴ ∠PQR = 45o => The bearing of P from R = 180 + 45 = 225o

Iron (soft iron) is usually preferred to steel in the making of electromagnets because it is more easily magnetized and equally more easily demagnetized than [II, III and IV]

Amphoteric oxide reacts with both acid and alkaline and only amphoteric oxide can displace hydrogen from alkaline solution (Al, Pb, Sn and Zn).

As global warming continues to exacerbate sea level rise and extreme weather, our nation's floodplains are expected to grow.

Benzene undergoes substitution reaction whereby one or more hydrogen atoms on the benzene ring is/are replaced by other atoms or groups of atoms. E.g CH\(_3\), OH, NO\(_2\), NH\(_2\) etc can be substituted to replace hydrogen atom. Substitution reaction is a reaction where one functional group is exchanged for another. There are several types of substitution reactions, including nucleophilic, electrophilic, and free radical. A substitution reaction is more peculiar to benzene compared to an addition reaction; benzene primarily undergoes substitution reactions because addition reactions would disrupt its stable aromatic ring structure, causing it to lose its unique stability due to delocalized electrons.

\theta = 30 x 2(∠ at center twice ∠ at circumference) ∴ \theta = 60^\circ Length of arc = \(\frac{\theta}{360}\times 2\pi r\) = \(\frac{60}{360}\times 2 \times \frac{22}{7}\times \frac{21}{1}\) = 22cm

Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R Mid point QR = \(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2}\) = \(\frac{2 + 0}{2}, \frac{1 + 0}{2}\) = \((1, \frac{1}{2})\) Gradient of QR = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) = \(\frac{1 - 0}{2 - 0}\) = \(\frac{1}{2}\) Gradient of PM = \(\frac{-1}{\frac{1}{2}}\) = -2 Equation of PM = y - y\(_1\) = m(x-x_\(_1\)) i.e y - 1/2 = -2(x-1) 2y - 1 = -4(x-1) 2y - 1 = -4x + 4 2y + 4x = 5

γ + S = 180(Allied angles) S = 180 - γ α + P = 180(Allied angles) p = 180 - α β = P + S β = 180 - α + 180 - γ β = 360 - α - γ β + α + γ = 360

\theta = 45 x 2 (∠ at center twice ∠ at \theta) \theta = 90^\circ Area of sector = \(\frac{\theta}{360}\) x \pir\(^2\) = \(\frac{90}{360}\) x \(\frac{22}{7}\) x 7 x 7 = \(\frac{77}{2}\)cm\(^2\) Area of triangle = \(\frac{1}{2}\) ab sin \theta = \(\frac{1}{2}\) x 7 x 7 sin 90 = \(\frac{1}{2}\) x 7 x 7 x 1 = \(\frac{49}{2}\) cm\(^{2}\) Area of shaded region = \(\frac{77}{2}\) - \(\frac{49}{2}\) = \(\frac{28}{2}\) = 14cm\(^2\)