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Pr(2) = 0.15 \(\frac{m}{32 + m + 25 + 40 + 28 + 45}\) = 0.15 \(\frac{m}{m + 170}\) = 0.15 m = 0.15(m + 170) m = 0.15m + 25.5 m - 0.15m = 25.5 \(\frac{0.85m}{0.85} = \frac{25.5}{0.85}\) m = 30 (b) Number of time = 32 + 30 + 25 + 40 + 28 + 45 = 200 times (c) Pr(even number) = \(\frac{30 + 40 + 45}{200}\) = \(\frac{115}{200}\) = \(\frac{23}{40}\)

x o\(^o\) 15\(^o\) 30\(^o\) 45\(^o\) 60\(^o\) 75\(^o\) 90\(^o\) 105\(^o\) 120\(^o\) 135\(^o\) \(^o\) y 0.0 1.5 2.6 2.1 2.6 1.5 0 - 1.5 - 2.6 - 3 -2.6 y = 35m\(^2\)x when x = 15\(^o\) y = 3 sin = (15\(^o\)) y = sin 30\(^o\) y = 3 x \(\frac{1}{2}\) = 1.5 When x = 30\(^o\) y = 3 sin 2(30\(^o\)) y = 3 sin 60\(^o\) = 2.6 When x = 45\(^o\) y = 3sin45\(^o\) = 2.1 When x = 60\(^o\) y = 3 sin 2(60\(^o\)) = 2 sin 120\(^o\) = 2.6 When x = 75\(^o\) y = 3 sin 2(75\(^o\)) = 3 sin 150\(^o\) = 1.5 When x = 90\(^o\) y = 3 sin 2(90\(^o\)) = 3 sin 180\(^o\) = 0 When x = 105\(^o\) y = 3 sin 2(105\(^o\)) = sin 210\(^o\) = -1.5 When x = 120\(^o\) y = 3 sin 2(120\(^o\)) = 3 sin 240\(^o\) = -2.6 When x = 135\(^o\) y = 3 sin 2(135\(^o\)) y = 3 sin 270\(^o\) y = -3 (b SEE THE GRAPH ABOVE... (c)(i) 113 (ii) 3\(^o\) and 82\(^o\)

Surface Area of the structure = curved surface area of cone + surface area of hemisphere = \(\pi rl + 2 \pi r^2\) But slant height l = l\(^2\) = 48\(^2\) + 14\(^2\) l\(^2\) = 2500 l = \(\sqrt{2500}\) = 50 Surface Area of the structure = \(\pi rl + 2 \pi r^2\) = \(\frac{22}{7}\) x 14 x 50 + 2 x \(\frac{22}{7}\) x 14 x 14 = 3432.0m\(^2\) ≈ 3430m\(^2\) to 3 s.f (b) Let Musah age be M Let Sesay be S M - 5 = 2(S - 5) ......(i) M + S = 100 ......(ii) From eqn ii) M = 100 - S By substitution, 100 - S - 5 = 2(S - 5) 95 - S = 2S - 10 3S = 105 \(\frac{3S}{3} = \frac{105}{3}\) S = 35 Therefore, Sesay's present age is 35 years

(a) Let the salary be x Shopping mall = \(\frac{1}{4}\) x \(x = \frac{x}{4}\) Open market = \(\frac{1}{3} \times x = \frac{x}{3}\) The remainder x - [\(\frac{x}{4} + \frac{x}{3}\)] = x - [\(\frac{3x + 4x}{12}\)] = \(\frac{x}{1} - \frac{7x}{12}\) = \(\frac{12x - 7x}{12}\) = \(\frac{5x}{12}\) mechanic = \(\frac{2}{5} \times \frac{5x}{12} = \frac{x}{6}\) \(\frac{x}{6} + \frac{x}{3} + \frac{x}{4}\) + 225,00 = x \(\frac{2x + 4x + 3x}{12}\) + 225,000 \(\frac{9x}{12}\) + 225,000 = x 225,000 = \(\frac{x}{1} = \frac{9x}{12}\) 225,000 = \(\frac{12x - 9x}{12}\) 225,000 = \(\frac{3x}{12}\) \(\frac{225,000 \times 12}{3}\) = x N900,000 = x (i) Her monthly income = N900,000 (ii) Amount in open market = \(\frac{N900,000}{3}\) = N300,000 (b) T\(_3\) = 4m - 2n T\(_9\) = 2m - 8n a + 2d = 4m - 2n .....(i) a + 8d = 2m - 8n ......(ii) a + 8d = 2m - 8n -(a + 2d = 4m - 2n \(\overline{6d = -2m - 6n}\) 6d = -2(m + 3n) d = -\(\frac{m + 3n}{3}\) Common difference = -\(\frac{m + 3n}{3}\)

(a) diagram above (b)(i) m\(^2\) = 5\(^2\) + 9\(^2\) m\(^2\) = 25 + 81 m\(^2\) = 106 m = \(\sqrt{106}\) = 10.296 km/hr Speed = \(\frac{distance}{times}\) distance = speed x time = 10.296 x 2 = 21km (b)(ii) tan\(\theta\) = \(\frac{9km/hr}{5km/hr}\) tan\(\theta\) = 1.8 \(\theta\) = tan\(^{-1}\) \(\theta\) = 60.9\(^o\) the bearing of cyclist x from y 90\(^o\) + 19.9\(^o\) = 109.9\(^o\) = 110\(^o\) (c) The average speed = \(\frac{distance }{time}\) = \(\frac{21}{4}\)km = 5km/her

Marks Upper Class Boundary Frequency (F) Cumulative Frequency (F) (CF) 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 7 11 17 20 29 34 30 25 21 6 7 18 35 55 84 118 148 173 194 200 (b) (c) (I) Median Mark = 54.5 (ii) Lowest mark, if 5% of the students passed = 14

(a) (i) < NPQ = 180\(^o\) - 50\(^o\) = 130\(^o\) < MNP = 130\(^o\) - 90\(^o\) = 40\(^o\) < NMP = 40\(^o\) (Isolceles triangle) < MNP = 180\(^o\) - (40\(^o\) + 40\(^o\)) = 180\(^o\) - 80\(^o\) = 100\(^o\) (ii) < OPQ = 90\(^o\)(50\(^o\) - 40\(^o\)) = 80\(^o\) < OPQ = 80\(^o\) (Isoseles triangle) < POQ = 180\(^o\) (80\(^o\) + 80\(^o\)) 180\(^o\) - 160\(^o\) = 20\(^o\) (b) 8y + 4x = 24 \(\frac{8y}{8} = \frac{-4y}{8} + \frac{24}{8}\) y = \(\frac{-1}{2}x + 6\) m = \(\frac{-1}{2}\) The equation will give \(\frac{y - 12}{x + 8} = \frac{-1}{2}\) 2y - 24 = x - 8 2y = - x - 8 + 24 2y = -x + 16

< AOB = 108\(^o\) - (90\(^o\) + 40\(^o\)) = 180\(^o\) - 130\(^o\) = 50\(^o\) < DCO = 40\(^o\) (Alternate angle) < ODE = \(\frac{180^o - 80^o}{2}\) = \(\frac{100^o}{2}\) = 50\(^o\) (b)(i) (b)(ii) Area of the parallelogram = ab sin \(\theta\) = 5cm x 7cm sin 125\(^3\) = 35 sin 125 = 28.7cm\(^2\) (c) x = \(\frac{1}{2}(1 - \sqrt{2})\) evaluate (2x\(^2\) - 2x) = 2[\(\frac{1}{2}(1 - \sqrt{2})^2\)] - 2[\(\frac{1}{2} (1 - \sqrt{2}\))] = 2[\(\frac{1}{4}(1 - 2\sqrt{2} + 2)\)] - 1 + \(\sqrt{2}\) = 2[\(\frac{1}{4}(3 - 2\sqrt{2})\) - 1 + \(\sqrt{2}\)] 2[\(\frac{3}{4} - \frac{1}{2} \sqrt{2} - 1 + \sqrt{2}\)] = \(\frac{3}{2} - \sqrt{2} - 2 + 2\sqrt{2}\) = \(\frac{1}{2} + \sqrt{2}\)

The part labeled I is commonly called tactile hairs, sinus hairs, whiskers, or vibrissae. ( are the long, specialized hairs found on the face and nose of rodents like rats, etc, and other animals like pussy cats)

specimen Test Observation time(sec) Medium of spread sense organ p 2 drops of specimen p + water 5 - 10 liquid / water Eye(s) Q Leveled teaspoonful of specimen Q + water 300 - 1800 liquid / water Eye(s) R Specimen R was cut open 0.1 - 2 Gas / Air Nose (b) (b) Biological Phenomenon that Occurred: Diffusion. (c) Factors that will Increase the Rate of Phenomenon for each Activities: (i)-Decreased concentration of the potassium permanganate: -Shaking/stirring/agitating: -Increased.temperature/heat. (ii)-Grinding/increasing the surface area of Specimen Q to reduce the molecular/grain size; -Decreased quantity of Specimen Q; Shaking/stirring/agitating;-Increased temperature/heat. (iii) -Increased air current/breeze/fan; Increased quantity of specimen R;-increased temperature/heat. Other types of fingerprints:-Whorl Loop;-Arch, Contour,- etc. (any other three types different from the One or the thumbprint). (V) Advantages ot Fingerprints: or crime detectton.securing documents.-Highly accurate/secure.-Unique/non-transferrable. -identification,-Data base.-Genetic engineering. Most economical technique/cost effective.Easy to use/ease of use. -Use ot smalt storage space/little storage space needed.-Provides accountability (biometric proof to show person was there when an event occurred). (V) Type of Variation: Discontinuous/Morphological.

Insulin helps control blood glucose levels by signaling the liver and muscle and fat cells to take in glucose from the blood.

Deamination is used to break down amino acids for energy, and occurs primarily in the liver.

The living components of an ecosystem are called the biotic components, option A has all living components.

Commensalism is a type of symbiosis, specifically, a biological relationship in which one specie benefits from an interaction, while the host species is neither positively or negatively affected to any tangible degree.

Hyphae that have walls between the cells are called septate hyphae; hyphae that lack walls and cell membranes between the cells are called nonseptate or coenocytic hyphae)