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Average performance rating of Team B = \(\frac{7+9+1+9+6}{5} = \frac{32}{5}\) = 6.4 Average performance rating of Team A = \(\frac{5+3+6+10+2}{5} = \frac{26}{5}\) = 5.2 ∴ The difference in the average performance ratings between Team B and Team A = 6.4 - 5.2 = 1.2

Total number of students that scored at most 50 marks = 100 + 80 + 60 + 40 + 80 = 360

SYLLABUS has 8 letters, 2S's and 2L's \(\frac{8!}{2! 2!}\)

If the majority are women implies that: 3 Women and 2 Men Or 4 Women and 1 Man = \(^4C_3 \times^6C_2+^4C_4\times^6C_1\) = \(4 \times 15 + 1 \times 6\) = 60 + 6 = 66

Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Total number of possible outcomes = 2 x 2 x 2 = 8 favourable number of outcomes = 7 ∴Pr(at least one head) = \(\frac {7}{8}\)

Let the number of white balls be n The number of red balls = 8 Now, the probability of drawing a white ball = \(\frac {n}{n + 8}\) The probability of drawing a red ball = \(\frac {8}{n + 8}\) Since Pr(White ball) = \(\frac{1}{2}\) x Pr(Red ball) \(\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}\) = \(\frac {n}{n + 8} = \frac {4}{n + 8}\) \(\therefore n = 4\) Number of white balls = 4 The possible number of outcomes = n + 8 = 4 + 8 = 12 Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball) Pr(Red ball) = \(\frac {8}{12}\) Pr(Red ball) = \(\frac {4}{11}\) Pr(Red ball and white ball) = \(\frac {8}{12} \times \frac {4}{11}\) \(\therefore\)Pr(Red ball and white ball) = \(\frac {8}{33}\)

Total possible outcome = 6 x 6 = 36 Required outcome = 15 ∴ Pr(E) = \(\frac{15}{36} = \frac{5}{12}\)

To find the point of intersection, equate the two equations ⇒ 2x\(^2\) + 10 = 4x + 16 ⇒ 2x\(^2\) + 10 - 4x - 16 = 0 ⇒ 2x\(^2\) - 4x - 6 = 0 Factorize ⇒ 2(x + 1)(x - 3) = 0 ∴ x = -1 or 3 The curves will intersect at -1 and 3 Area = \(\int^b_a\) [upper function] - [lower function] dx = \(\int^3_1 4x + 16 - (2x^2 + 10) dx\) = \(\int^3_1 -2x^2 + 4x + 6dx\) = \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\) = \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\) = 18 - \((\frac {10}{3})\) = 18 + \(\frac {10}{3}\) = \(\frac {64}{3}\)

The patient cannot see clearly an object closer than 50 cm Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens u=25cm ;v=-50cm (virtual image); p= \(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\) ⇒\(\frac{1}{f}\) =\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = f = 50cm =0.5m ⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\) p = \(\frac{1}{f}\) p = \(\frac{1}{0.5}\) p = 2 diopters ; The patient needs a converging lens with a power of 2 diopters

Mechanics is the branch of physics that deals with the motion of objects and the forces acting on them. It is the foundation of all other branches of physics, and it is used to study everything from the motion of planets to the motion of electrons.

m=1200kg, \(V_1\)= \(10ms^{-1}\) \(V_2\) = \(15ms^{-1}\), w= ? work=►K.E = \( K.E_2\) = \(K.E_2\) - \(K.E_1\) ⇒work= \(\frac{1}{2}{mv^2_2}-\frac{1}{2}{mv^2_1}\) ⇒work= \(\frac{1}{2}m({v^2_2}-{v^2_1}\)) ⇒work= \(\frac{1}{2}\times 1200\times (15^2-10^2)\) ⇒work = 600 \times (225 -100) ⇒work= 600 \times 125 ⇒work= 7.5\times\(10^4 j\)

The work done against the gravitational force is calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately \(9.8 ms^2\) on Earth), and h is the height. Substituting the given values, we get W = 3.0 kg * \(9.8 m/s^2\) * 240 m = 7.2 kJ. Therefore, the correct answer is '7.2 kJ'.

ε= 80%, L=200\times 10 = 2000N, \(d_L\)=10m \(d_E\)= 110-10 = 100m, E=? ε = \(\frac{work done on the load}{work done by the effort}\) \times 100% ⇒ work done on load = 2000 \times10 =20,000 j ⇒ work done by effort = E \times 100 = 100E ⇒ 80 = \(\frac{20,000}{100E}\) \times 100% ⇒ 80 = \(\frac{20,000}{E}\) ⇒ E = \(\frac{20,000}{80}\) ⇒ E = 250N

We first find the lorry's initial velocity a = \(4ms^-2\), s= 250m, t= 10, u=? ⇒ ut + \(\frac{1}{2}\times4\times(10^2) ⇒250= 10u + 200 ⇒ 250-200 = 10u ⇒ 50 = 10u ⇒U =\frac{50}{10}\) = \(5ms^-1\) For the next 10s, we set t=20s = s = (5 \times 20) + \(\frac{1}{2}\times4\times{20^2}\) = s = 100 + 800 ⇒ 900 m ;Distance covered:900-250 =650m

W = mg = 220 x 9.8 = 2156 N ⇒Sin 38º = \(\frac{2156}{T_1}\) ⇒ \(T_1\) = \(\frac{2156}{Sin 38}\) ⇒ \(T_1\) = 3502 N Cos 38º = \(\frac{T_2}{T_1}\) ⇒ \(T_2\) = 3502 x Cos 38º ⇒ \(T_2\) = 2760 N ; \(T_1\) = 3502 N, \(T_2\) = 2760 N.