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a. solution above bi. From ∆PMT the height of the building is PT \(Tan\theta = \frac{opposite}{adjacent}\) tan66^\circ = \(\frac{PT}{50}\) Therefore, PT = \(tan 66^\circ\times50 = 112.3m\) ( to 1 d.p) bii. From ∆PCT \(Tan53^\circ = \frac{PT}{CT}\) \(Tan53^\circ = \frac{112.3}{CT}\) \(Tan53^\circ\times CT = 112.3\) \(CT = \frac{112.3}{Tan53^\circ}\) = 84. 62m. MC = CT - MT = 84.62 - 50 = 34.6m Therefore, MC = 34.6m ( to 1 d.p ).

Consider |NS| \theta = 180^\circ - (90^\circ + 35^\circ) (sum of angles on a straight line is 180^\circ) = 180^\circ - 90^\circ - 35^\circ = 90^\circ - 35^\circ = 55^\circ ∴ His new bearing is N55^\circE.

a. the diagram above solution to question a. bi. Using the sine rule \(\frac{sinP}{XY} = \frac{sinY}{PX}\) = \(\frac{sin93^\circ}{180} = \frac{sinY}{60}\) cross multiply \(sinY\times180 = sin93^\circ\times60\) sinY = 0.3329 Y = \(sin^{-1}0.3329\) Y = 19.44^\circ the angle between 30^\circ and 19.44^\circ at Y = 30^\circ - 19.44^\circ = 10.56^\circ Therefore, the bearing X from Y = 360^\circ - 10.56^\circ = 349^\circ( to 3 sig. figure) bii. Using cosine rule Let a = |PY|, b = |PX| and c = |XY| \(a^2 = b^2 + c^2 - 2bcCos A\) where A = 67.56^\circ \(a^2 = 60^2 + 180^2 - 2(60)(180)cos 67.56^0\) \(a2 = 3600 + 32400 - 8245.06\) \(a^2 = 27754.94^0\) a = \(\sqrt{27754.94}\) a = 166.6 m ∴ |PY| = 167 m (to 3 s.f)

Total number of students = total frequency = 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 Since the failed mark was 4, Total number of students that passed = 12 + 7 + 5 + 10 = 34 Pr( passed) = \(\frac{34}{53}\) = 0. 64

Number of students that scored at least 3 marks = 9 + 1 + 12 + 7 + 5 + 10 = 44

x̄ = \(\frac{ 14 + 15 + 16 +17 + 18 + 19}{6} = \frac{99}{6}\) = 16.5 M . D = \(\frac{ (14 - 16.5) + (15 - 16.5) + (16 - 16.5) + (17 - 16.5) + (18 - 16. 5) + (19 - 16.5) }{ 6}\) M.D = \(\frac{(-2.5) + (-1.5) + (-0.5) + (0.5) + (1.5) + (2.5)}{6}\) Taking the absolute value of the deviations M.D = \(\frac{2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5}{6}\) M.D = \(\frac{9}{6}\) = 1.5

a. X F FX 3 4 12 4 6 24 5 4 20 6 5 30 7 6 42 8 2 16 Σf =27 Σfx = 144 Median position = \(\frac{∑F + 1}{2}\) (since ∑F is odd so we added 1) \(\frac{27 + 1}{2} = \frac{28}{2}\) = 14th position. counting down the frequency column, the median position occurs at 4. Therefore, the median height is 5m. b i. mean(x̄) = \(\frac{Σfx}{Σf}\) = \(\frac{144}{27}\) ∴ x̄ = 5.3m ( to 1 d.p) ii. x f fx \(fx^2\) 3 4 12 (3 x 3)4 =36 4 6 24 (4 x 4)6 =96 5 4 20 ( 5 x 5)4 =100 6 5 30 (6 x 6)5 =180 7 6 42 (7 x 7)6 =294 8 2 16 (8 x 8)2 =128 Σf =27 Σfx = 144 Σ\(fx^2\) = 834 Standard deviation = \(\sqrt{vaiance} = \sqrt{\frac{Σfx^2}{Σf} - (\frac{Σfx}{Σf})^2}\) S D = \(\sqrt{\frac{834}{27} - (\frac{144}{27})^2}\) S.D = \(\sqrt{30.888 - (5.333)^2}\) = \(\sqrt{30.888 - 28.441}\) = \(\sqrt{2.447}\) = 1.5738 ≈ 1.6 ( to 1 d.p)

Total numbers from 40 to 50 inclusive = 11 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50). Prime numbers between 40 and 50 (inclusive) are 41, 43, and 47. There are 3 prime numbers in this range. therefore, Pr(prime numbers) = \(\frac{3}{11}\)

The magnitude of the electric field strength at any point between the plates is not directly proportional to the magnitude of the charge on each plate. Instead, it is directly proportional to the charge density (charge per unit area) on the plates. Therefore, if the charge on the plates is increased while keeping the area of the plates constant, the electric field strength will increase. However, if the area of the plates is also increased such that the charge density remains constant, the electric field strength will not change.

Number of cycles, n = \(2\frac{1}{2} = \frac{5}{2}\) Length = n\lambda \lambda = \(\frac{Length}{n}\) = \(\frac{125}{\frac{5}{2}}\) Therefore, \lambda = 50cm = 0.50m.

F = qvB ( force on moving charge in a magnetic field) F = \(\frac{mv^2}{r}\) equating F qvB = \(\frac{mv^2}{r}\) qB = \(\frac{mv}{r}\) qBr = mv Therefore, v = \(\frac{qBr}{m}\)

The pitch of a sound is determined by its frequency. Higher frequencies correspond to higher pitches. For example, a high-pitched sound like a whistle has a higher frequency than a low-pitched sound like a drum.

The expression for the escape velocity \(v_e\) of an object from a planet of mass M and radius R using the universal gravitational constant G is: \(v_e = \frac{\sqrt{2GM}}{R}\)

An atom in an excited state is one whose electrons have moved to a higher energy level. In an atom, electrons are typically found in specific energy levels or orbitals. When an electron absorbs energy, it can move from a lower energy level to a higher energy level, which is referred to as an excited state. In this higher energy state, the electron is farther from the nucleus and has more potential energy. When the electron returns to its lower energy level, it emits the excess energy as electromagnetic radiation, often in the form of visible light. This is what gives rise to the characteristic spectral lines observed in spectroscopy.

V = 220V, P = 1.60KW = 1060W, I = ? P = IV I = \(\frac{P}{V} = \frac{1060}{220}\) ≈ 5A.