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Volume of prism = Area x height In \(\Delta\) RQP, QR\(^2\) + QP\(^2\) = RP\(^2\) 3\(^2\) + QP\(^2\) = 5\(^2\) QP\(^2\) = 5\(^2\) - 3\(^2\) QP = \(\sqrt{16}\) = 4 cm \(\therefore\) Area = \(\frac{1}{2} \times base \times height\) = \(\frac{1}{2} \times 3 \times 4\) = 6 cm\(^2\) Volume = 6 x 7 = 42 cm\(^3\)

S = \(2\pi r^2 + 2\pi rh\) = \(2\pi r(r + h)\) = \(2\pi (7) (7 + 10)\) = \(238\pi cm^2\)

Area of trapezium = \(\frac{1}{2}\)( a + b) x h a = 9cm, b = 15cm, c = (15 - 9) = 6cm since height was not given, so we calculate for it using trig . ratio h = 2\(\sqrt{3}\)cm Area of trapezium = \(\frac{1}{2}\)( a + b ) x h = \(\frac{1}{2}\)( 9 + 15 ) x 2\(\sqrt{3}\) = 24 x \(\sqrt{3}\) = 24\(\sqrt{3}\)cm\(^2\)

S.D = \(\sqrt{Variance}\) = \(\sqrt{25}\) = 5

U = {a, b, c, d, e, f, g} P = {b, d, e, f} Q = {a, c, f, g} P' = {a, c, g} P' \(\cap\) Q = {a, c, g}

\(\bar{RP} = 3 - (-2)\) = 5

< T = \(\frac{x}{1}\) = 25o (PQ = QT) < SQR = 2(25o) = 50o (sum of interior angle) < Q + < R + < S = 180o 50o + 75o + < S = 180o = 125o + < S = 180o < S = 180o - 125o = 55o

Using cosine formula (t\(\sqrt{3}\))2 = t2 + t2 - 2t2 cos\(\theta\) 3t2 = 2t2 - 2t2 cos\(\theta\) = 2t2(1 - cos\(\theta\)) 1 - cos\(\theta\) = \(\frac{3t^2}{2t^2}\) = \(\frac{3}{2}\) cos = 1 - \(\frac{3}{2} = -\frac{1}{2}\) \(\theta\) = cos-1(-\(\frac{1}{2}\)) = 120o and 240o N.B 0 \(\geq\) \(\theta\) 360

\(\begin{array}{c|c} \text{colour of cars} & \text{Number (frequency)} \\ \hline yellow & 3 \\white & 4\\ red & 8\\ green & 2\\ blue & 6\\ black & 2\\ \hline & 25 \\ \hline\end{array}\) Thus, the fraction of the total numbers that are yellow is \(\frac{3}{25}\)

The interquartile range formula is the first quartile subtracted from the third quartile: IQR = Q3 – Q1.

\(\begin{array}{c|c} \text{no. of passengers} & \text{Number of taxis}\\ \hline 0.5 - 2.5 & 3\\ 2.5 - 4.5 & 4 \\ 4.5 - 6.5 & 7\\ 6.5 - 8.5 & 5\\ 8.5 - 10.5 & 4 \\ 10.5 - 12.5 & 1\\ \hline \text{Total} & 24 \end{array}\) Thus, the taxi with more than 4 passengers = 7 + 5 + 4 + 1 = 17

s = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\) s\(^2\) = \(\frac{a^2}{x^2} - \frac{b^2}{y^2}\) \(\frac{b^2}{y^2}\) = \(\frac{a^2}{x^2}\) - s\(^2\) \(\frac{b^2}{y^2}\) = \(\frac{a^2 - s^2 x^2}{x^2}\) \(\frac{1}{y^2}\) = \((\frac{a^2 - s^2 x^2}{x^2}) \times \frac{1}{b^2}\) \(\frac{1}{y^2}\) = \(\frac{a^2 - s^2 x^2}{b^2 x^2}\) \(\therefore\) y\(^2\) = \(\frac{b^2 x^2}{a^2 - s^2 x^2}\) y = \(\sqrt{\frac{b^2 x^2}{a^2 - s^2 x^2}}\)

Using these: Loss = C.p - S.p : Percentage Loss = ( Loss X 100) \div C.p =15 %

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

From the diagram above, the bearing of X from Y is 330º (The bearing from Y to X is the opposite direction of the bearing from X to Y. Since bearings are measured clockwise, to find the reverse bearing, you need to add 180^\circ to the original bearing and then adjust for any values over 360^\circ)