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Number of letter in MATHEMATICS = 11 Number of letter M = 2 Number of letter T = 2 Number of letter A = 2 Arrangement = 11!/(2! 2! 2!) ways

Angle corresponding to 4 is 40/240 x 360/1 = 60^\circ Note that total angle in a circle is 360^\circ. Note also that sum of the frequencies given is 240.

U = {1, 2, 3, 4, 5,..., 20} E1 = {3, 6, 9, 12, 15, 18} E2 = {4, 8, 12, 16, 20} P(E\(_2\)) = \(\frac{5}{20}\) P(not E\(_2\)) = 1 - (\(\frac{5}{20}\)) = \(\frac{15}{20} = \frac{3}{4}\)

Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation). Sum of (x) = 15x, algebraic sum of (deviation) = zero (0) Sum of (squared-deviation) = 10x\(^2\) Variance = \(\frac{∑(squared-deviation)}{\text{n}}\) = \(\frac{10x^2}{5}\) = 2x\(^2\)

Range = Highest - lowest number => 10-4 = 6 Mode is the number with highest occurrence => Mode = 10 Sum = 6 + 10 = 16

No of ways of choosing 1 man, 2 women = 5C1 x 3C2 No of ways of choosing 2 men, 1 woman = 5C2 x 3C1 Summing, => (5C1 x 3C2) + (5C2 x 3C1) = 15 + 30 = 45

Hints: 1. Integrate the given first derivative of f(x) at the boundaries, (0,0) Then solve accordingly to get f(x) = y = \(\frac{3x^2}{2}\) + 2x

At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5; f(1) => a + b + c = 5 .....(1) Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2) From (2),=> b = 1, and 2ax = 2x, => a = 1. ( comparing coefficients) substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3 Thus a = 1, b = 1 and c = 3

\Delta PQR is isosceles ∴∠PRQ = 30o Also in \Delta PQR ^Q = 180 -(30 + 30) = 120o PQRS is a cyclic quad ∴^S + ^Q = 180(opp ∠s of a cyclic quad) S + 120 = 180 S = 60o In \DeltaPSR x + 50 + 60 = 180(sum of ∠s of a \Delta) x + 110 = 180 x = 180 - 110 x = 70o

In the given diagram, < PQS = \(\frac{1}{2}\) x 96º = 48º ( Therefore, < PST = 48º ( alternate segment).

104\(_x\) = 68 1 \(\times x^2 + 0 \times x + 4 \times x^0\) = 68 x\(^2\) = 68 - 4; x\(^2\) = 64 x = \(\sqrt{64}\) = 8 (104\(_x\) change to base 10)

8 71 8 8 rem 7 8 1 rem 0 0 rem 1 71\(_8\) = 107\(_8\)

5-3log52 x 22log23 i Let -3log52 = p => log52-3 = p ∴2-3 = 5p ∴5-3log52 = 5log52-3 = 5p ii 22log23 = q => log532 = q ∴32 = 2q ∴222log23 = 2q = 32 = 9 i x ii = 2-3 x 9 = 1/23 x 9 = 1/8 x 9 = 9/8 = 11/8

\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\) = \(\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}\) = \(\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\) = \(\frac{27}{8}\times \frac{10}{9} = \frac{15}{4}\) OR We start with the expression: \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \sqrt{\frac{100}{81}}\) First, simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\): \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left(\frac{81}{16}\right)^{\frac{3}{4}} = \frac{81^{\frac{3}{4}}}{16^{\frac{3}{4}}}\) Calculating \(81^{\frac{3}{4}}\) and \(16^{\frac{3}{4}}\): \(81 = 3^4 \implies 81^{\frac{3}{4}} = 3^3\) = 27 \(16 = 4^2 \implies 16^{\frac{3}{4}} = 4^{\frac{3}{2}} = 2^3\) = 8 Thus, \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \frac{27}{8}\) Next, simplify \(\sqrt{\frac{100}{81}}\): \(\sqrt{\frac{100}{81}} = \frac{\sqrt{100}}{\sqrt{81}} = \frac{10}{9}\) Now, multiply the two results: \(\frac{27}{8} \times \frac{10}{9} = \frac{27 \times 10}{8 \times 9} = \frac{270}{72}\) Finding the GCD of 270 and 72 gives: \(\frac{270 \div 18}{72 \div 18} = \frac{15}{4}\) The final result is: \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \sqrt{\frac{100}{81}} = \frac{15}{4}\)

3y\(^2\) = 27y 3y\(^2\) - 27y = 0 3y(y - 9) = 0 3y = 0 or y - 9 = 0 y = 0 or y = 9 y = 0 or 9