Explanation

(a) Authority is of two kinds: Institutional and functional: Institutional Authority is a group of people or individual who legitimately occupies Public Office with capacity to determine public goals and policies and direct the execution or otherwise of such goals and policies e.g. School Principal, Authority of Head of State, Native Authority, Judicial Authority while Functional authority is the power which accrues to a group of people or an individual as a result of the position he holds in an organization. It is the power which is lawful and conforms with accepted social norms. It is a legitimate right to direct or command both human and other resources.

(b) Main sources of Political Authority are:

(i) Tradition or custom e.g. Monarchy / Chieftaincy. This is common in traditional society.

(ii) Legality or law e.g. Constitution. This is the case in modern society.

(iii) Charisma or exceptional personal character of the leader and a good example is Hiltler.

(iv) Consent of the people. People may decide to invest a certain person with authority.

Correct Answer: A. 7

Explanation

The median mark is the middle mark when the marks are arranged in ascending order. First, let's arrange the marks in ascending order: 5, 6, 7, 7, 7, 8, 8, 9, 9

Since there are 9 students, the median mark is the average of the 4th and 5th marks. The 4th mark is 7 and the 5th mark is also 7, so the median mark is 7.

Therefore, the answer is 7.

Correct Answer: D. Austin Okocha

Correct Answer: B. Abidjan

Correct Answer: C. 1.5×10²⁴

Correct Answer: A. deliberate government policy and market

Correct Answer: A. 18.57NC−1

Explanation

nC\(_4\) = \(\frac{n!}{[n-4]! 4!}\) → \(\frac{n[n-1][n-2][n-3][n-4]!}{[n-4]!4!}\)

nC\(_5\) = \(\frac{n!}{[n-5]! 5!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5]!}{[n-5]!5!}\)

and nC\(_6\) = \(\frac{n!}{[n-6]! 6!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5][n-6]!}{[n-6]!6!]}\)

d = U2 - U1 = U3 - U2

nC\(_5\) - nC\(_4\) = \(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\)

nC\(_6\) - nC\(_5\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

nC\(_5\) - nC\(_4\) = nC\(_6\) - nC\(_5\)

\(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

divide both sides by n(n-1)(n-2)(n-3)

\(\frac{n-9}{5!}\) = \(\frac{[n-4][n-11]}{5! 6}\)

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)

6n - 54 = n\(^2\) -11n - 4n + 44

6n - 54 = n\(^2\) - 15n + 44

n\(^2\) - 21n + 98 = 0

n\(^2\) - 7n - 14n + 98 = 0

n(n - 7) -14(n - 7) = 0

(n-7)(n-14) = 0

n = 7 or 14

ii.

d = U2 - U1

when n = 7

d = 7C\(_5\) - 7C\(_4\)

d = \(\frac{7*6*5!}{2! * 5!}\) - \(\frac{7*6*5*4!}{3! * 4!}\)

d = 21 - 35

d = -14

when n = 14

d = 14C\(_5\) - 14C\(_4\)

d = \(\frac{14*13*12*11*10*9!}{9! * 5!}\) - \(\frac{14*13*12*11*10!}{10! * 4!}\)

d = 2002 - 1001

d = 1001

Correct Answer: (a) one hundred and one

Correct Answer: B. ionization energy