Correct Answer: D. \(76^0\)

Explanation

Reflex ∠MOR = 2 × 124° = 248° (angle at the centre is twice the angle at the circumference)

∠MOR = 360° - 248° = 112° (sum of angle at a point is 360°)

∠OMN = 360° - (124°+ 48° + 112°) (sum of angles in a quadrilateral is 360°)

= 360° - 284°

∴ ∠OMN = 76°

Explanation

(a) Factors responsible for population distribution in West Africa:

(i) Relief (ii) Security reasons

(iii) Mining (iv) Drainage. (v) Historical factors (vi) Agricultural/Fertile soil.

(vii) Presence of higher educational institution

(viii) Presence of higher educational institutions.

(ix) Political reasons

(x) Diseases/Pests

(xi) Cultural factor.

(xiii) Availability of water.

(xiv) Administration.

(xv) Presence if social amenities.

(xvi) Accessibility

(xvii) Commercial activities.

(b) Advantages of low population density.

(i) Abundant resources

(ii) Low crime rate/easy detection of crime

(iii) Low environment pollution.

(iv) High standard of living

(v) High per capita income

(vi) Social cohesion.

(vii) Low pressure on social amenities,

(viii) Easy flow of traffic

(ix) Prices of agricultural commodities are relatively low/Low cost of living.

Correct Answer: D. Cations are discharged.

Explanation

Electrolysis is the process of using an electric current to cause a chemical reaction to take place. The cathode is the negative electrode in an electrolytic cell. It is where cations (positively charged ions) are discharged

Correct Answer: C. is an intimate form of expression

Correct Answer: D. she is dead

Correct Answer: D. referendum

Correct Answer: A. Hydrogen and oxygen gases

Correct Answer: A. mucuna

Correct Answer: C. Sisera

Explanation

(a) 4 boys, 5 girls

P(both are boys) = P(1st is a boy and 2nd is a boy)

= \(\frac{4}{9} \times \frac{3}{8}\)

= \(\frac{1}{6}\)

(b) If p is the probability of a standard transistors, then \(p = 0.8 = \frac{4}{5}\)

If q is the probability of a transistor below standard, \(q = 0.2 = \frac{1}{5}\)

For 6 transistors, we have

\((p + q)^{6} = p^{6} + 6p^{5}q + 15p^{4}q^{2} + 20p^{3}q^{3} + 15p^{2}q^{4} + 6pq^{5} + q^{6}\)

(i) p(exactly 2 standard transistors) = \(15p^{2}q^{4}\)

= \(15 \times (\frac{4}{5})^{2} \times (\frac{1}{5})^{4}\)

= \(\frac{48}{3125}\)

(ii) p(exactly 1 standard transistor) = \(6pq^{5}\)

= \(6 \times \frac{4}{5} \times (\frac{1}{5})^{5}\)

= \(\frac{24}{15625}\)

(iii) p(at least 2 standard transistors) = 1 - [p(0 standard transistor) + p(1 standard transistor)]

p(0 standard transistor) = \(q^{6}\)

= \((\frac{1}{5})^{6} = \frac{1}{15625}\)

p(at least 2 standard transistors) = \(1 - [\frac{1}{15625} + \frac{24}{15625}]\)

= \(\frac{15600}{15625} = \frac{624}{625}\)

(iv) p(at most 2 standard transistors) = p(0 standard transistor) + p(1 standard transistor) + p(2 standard transistors)

= \(\frac{1}{15625} + \frac{24}{15625} + \frac{48}{3125}\)

= \(\frac{265}{15625} = \frac{53}{3125}\)