JAMB Mathematics Past Questions & Answers - Page 57

281.

The locus of a point which moves so that it is equidistant from two intersecting straight lines is the

A.

bisector of the two lines

B.

line parallel to the two lines

C.

angle bisector of the two lines

D.

perpendicular bisector of the two lines

Correct answer is C

The locus of a points equidistant from two intersecting straight lines is a pair of bisectors that bisect the angles formed by the two intersecting lines.

282.

Score (x) 0 1 2 3 4 5 6
Freq (f) 5 7 3 7 11 6 7

Find the mean of the data.

A.

3.26

B.

4.91

C.

6.57

D.

3.0

Correct answer is A

Score (x) 0 1 2 3 4 5 6   Freq (f) 5 7 3 7 11 6 7 46 fx 0 7 6 21 44 30 42 150

Mean = \(\frac{\sum fx}{\sum f}\)

= \(\frac{150}{46}\)

= 3.26

283.

Given matrix M = \(\begin{vmatrix} -2 & 0 & 4 \\ 0 & -1 & 6 \\ 5 & 6 & 3 \end{vmatrix}\), find \(M^{T} + 2M\)

A.

\(\begin{vmatrix} -4 & 2 & 1\\ 6 & 0 & 5 \\ 0 & 6 & 2 \end{vmatrix}\)

B.

\(\begin{vmatrix} -6 & 0 & 13\\ 0 & -3 & 18 \\ 14 & 18 & 9 \end{vmatrix}\)

C.

\(\begin{vmatrix} 5 & 2 & 6 \\ 0 & 1 & 1\\ 3 & 4 & -7 \end{vmatrix}\)

D.

\(\begin{vmatrix} -4 & 0 & 8 \\ 0 & -2 & -16 \\ 10 & 12 & 6 \end{vmatrix}\)

Correct answer is B

M = \(\begin{vmatrix} -2 & 0 & 4 \\ 0 & -1 & 6 \\ 5 & 6 & 3 \end{vmatrix}\)

M\(^{T}\) = \(\begin{vmatrix} -2 & 0 & 5 \\ 0 & -1 & 6\\ 4 & 6 & 3 \end{vmatrix}\)

2M = \(\begin{vmatrix} -4 & 0 & 8\\ 0 & -2 & 12\\ 10 & 12 & 6\end{vmatrix}\)

M\(^T\) + 2M = \(\begin{vmatrix} -6 & 0 & 13 \\ 0 & -3 & 18 \\ 14 & 18 & 9 \end{vmatrix}\)

284.

In how many ways can the word MATHEMATICIAN be arranged?

A.

6794800 ways

B.

2664910 ways

C.

6227020800 ways

D.

129729600 ways

Correct answer is D

MATHEMATICIAN = 13 letters with 2M, 3A, 2T, 2I.

Hence, the word MATHEMATICIAN can be arranged in \(\frac{13!}{2! 3! 2! 2!}\)

= 129729600 ways

285.

If a fair coin is tossed 3 times, what is the probability of getting at least two heads?

A.

\(\frac{2}{3}\)

B.

\(\frac{4}{5}\)

C.

\(\frac{2}{5}\)

D.

\(\frac{1}{2}\)

Correct answer is D

The outcomes are {HHH, HHT, HTT, HTH, THH, THT, TTH, TTT}

P(at least two heads) = \(\frac{4}{8}\)

= \(\frac{1}{2}\)