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Rearrange data in increasing order: 6,7,7,7,8,8,8,9,9 and 11 First quartile (ungrouped data) = \(\frac{n + 1}{4}\)th value = \(\frac{11}{4}\) = 2.75th value = \(\frac{7 + 7}{2}\) = 7.0

∑fx = (160 * 4) + (161 * 6) + (162 * 3) + (163 * 7) + (164 * 8) + (165 * 9) =640 + 966 + 486 + 1,141 + 1,312 + 1,485 = 6,030 ∑f = 4 + 6 + 3 + 7 + 8 + 9 = 37 = \(\frac{∑fx}{∑f}\) = \(\frac{6030}{37}\) = 162.97 or 163

Total angles = 360^\circ 100 + 60 + 80 + M = 360 M = 360 - 240 = 120 Orange (M) = 120^\circ and F ( fruits in total) For Apples: \(\frac{80}{360} \times F\) = 60 F = \(\frac{60 \times 360}{80}\) F = 270 fruits in total For Orange = \(\frac{120}{360} \times 270\) = 90 oranges

Given that 4, 16, 30, 20, 10, 14 and 26 Adding up = 120 nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92 The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\) = 276o

Given from 41 to 56 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56 The nos multiple of 9 are: 45, 54 P(multiple of 9) = \(\frac{2}{16}\) = \(\frac{1}{8}\)

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table above, event space, n(E) = 6 sample space, n(S) = 36 Hence, probability sum of scores is at least 10, is; \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

mc\(\Delta\theta\) = m\(\lambda\)\(_f\) 2 x 400 (100 - 0) = m(3.3 x 10\(^{5}\)) m = \(\frac{2 \times 400 \times 100}{3.3 \times 10^{5}} = \frac{8}{33}\)kg

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

Given m = 10\(^{-2}\) w = 100rads\(^{-1}\) F = mw\(^{2}\)r r = 0.2m F = ? = 10\(^{-2}\) x (100)\(^{2}\) x 0.2 = 20N

Simple harmonic motion (SHM) is a special type of periodic motion where the restoring force on the moving object is directly proportional to the magnitude of the object's displacement and acts towards the object's equilibrium position.

Givn m = 0.1kg u = 10ms\(^{-1}\) h = 10m g = 10ms\(^{-2}\) At h\(_{max}\); v = 0 Using \(\not v\) = \(\not u\) = \(\not gt\) \(\not 0\) = \(\not w\) - \(\not 10t\) v\(^{2}\) = u\(^{2}\) - 2gh 0\(^{2}\) = 10\(^{2}\) - 2 x 10h 20h = 100 h = 5m Total height = 10 + 5 = 15m Total energy = mghr = 0.1 x 10 x1 5 = 15J

power = \(\frac{force * distance}{time}\) → \(\frac{200 * 9}{5}\) power = 36w

\(\frac{\bigtriangleup s}{\bigtriangleup t}\) = uniform velocity

Given \(\frac{m_{1}}{m_{2}} = \frac{3}{1}\) \(\Rightarrow\) \(m_1\) = 3\(m_2\) \(\frac{a_{1}}{a_{2}} = \frac{2}{9}\) = \(a_1\) = \(\frac{2a_{2}}{9}\) \(\frac{F_{1}}{F_{2}}\) = ? Since F = ma \(\Rightarrow\) \(\frac{F_{1}}{F_{2}} = \frac{m_{1}a_{1}}{m_{2}a_{2}}\) = \(\frac{(3 \not m_{2})(2 \not a_{2})}{\not m_{2} a_{2} \not a_{3}}\) \(\frac{F_{1}}{F_{2}} = \frac{2}{3}\) \(\Rightarrow\) 2:3

Distance covered = Area under the graph D = area of square + area of trapezium D = L x L + 1/2(a+b)h D = 4 x 4 + 1/2( 4+ 10)8 D = 16 + 1/2 x 14 x 8 D = 16 + 56 D = 72m.