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\(\begin{array}{c|c} x & x - x & (x - x)^2\\ \hline 7 & -3 & 9\\8 & -2 & 4 \\9 & -1 & 1\\10 & 0 & 1\\11 & 1 & 1\\ 12 & 2 & 4\\13 & 3 & 9\\ \hline & & 28\end{array}\) S.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\) = \(\sqrt{\frac{\sum d^2}{N}}\) = \(\sqrt{\frac{28}{7}}\) = \(\sqrt{4}\) = 2

Chance of x = \(\frac{1}{2}\) Change of Y = \(\frac{2}{3}\) Chance of Z = \(\frac{1}{4}\) Chance of Y and Z only occurring = Pr (Y ∩ Z ∩ Xc) where Xc = 1 - Pr(X) 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\) = Pr(Y) x Pr(Z) x Pr(Xc) = \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\) = \(\frac{1}{12}\)

The two equations are y = x(2 - x) and y = (x - 1)(x - 3) The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0 put y = 0; 0 = x(2 - x) 0 = (x - 1)(x - 3) x = 0 or x = 2; x = 1 or x = 3 The values of x at P, Q, R are increasing towards the positive direction of x-axis at P, x = 1 at Q, x = 2 at R, x = 3 P, Q, R are respectively (1, 2, 3)

The total angle in a regular pentagon = \((2(5) - 4) \times 90\) = \(6 \times 90 = 540^\circ\) Each interior angle = \(\frac{540}{5} = 108^\circ\) While the interior of a quadrilateral = \(\frac{360}{4} = 90^\circ\) \(PTU + TPU + TUP = 180^\circ\) \(108^\circ + (180^\circ - 90^\circ) + TUP = 180^\circ\) \(TUP = 180^\circ - (108^\circ + 18^\circ) = 54^\circ\)

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90^\circ < RTS = < PQS = 32^\circ (Alternative angle) < PSQ = 90 - < PSQ = 90^\circ - 32^\circ = 58^\circ

OQ = OR = radii < ROQ = 180 - 96 = 84º \(\bigtriangleup\)OQR = Isosceles R = Q R + Q + 84 = 180(angle in a \(\bigtriangleup\)) 2R = 96 since R = Q R = 48º ORQ = 48º

Construction joins R to P such that SRP = straight line R = 180o - 107o < p = 180o - (107o - 34o) 108 - 141o = 39o Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT < S = < T = 39o SRT = 180 - (39o + 39o) = 180o - 78o = 102o

Since QRU= Xo RSU = Xo, But RSU = 180o - (50o + 35o) = 180o - 85o = 95o x = 95o

Using angle bisector theorem: line PS bisects angle QPR QS/QP = SR/PR 3/4 = SR/9 4SR = 27 SR = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)cm

SQ2 + OS2 = OQ2 + 52 = 142 SQ2 = 142 - 52 196 - 25 = 171 ST2 + TQ2 = SQ2 ST2 + 22 = 171 ST2 = 171 - 4 = 167 ST = \(\sqrt{167}\) = 12.92 = 12.9cm

Since area of square PQRS = 100cm2 each lenght = 10cm Also TUYS : XQVU = 1 : 16 lengths are in ratio 1 : 4, hence TU : UV = 1: 4 Let TU = x UV = 1: 4 hence TV = x + 4x = 5x = 10cm x = 2cm TU = 2cm UV = 8cm But TU = SY and UV = YR YR = 8cm

Number f students that took the test = \(\sum f\). Where f is the frequencies = 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.