Mathematics 2019 Jamb Examination

The sample space for 3 tosses is \( \displaystyle 2^3 = 8 \):
{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. '
At least two heads' means 2 or 3 heads. The favorable outcomes are {HHH, HHT, HTH, THH}. \[ \displaystyle \text{Probability} = \frac{4}{8} = \frac{1}{2} \]

The word MATHEMATICIAN has 13 letters. We count the repetitions: - M: 2 times - A: 3 times - T: 2 times - I: 2 times \[ \displaystyle \text{Arrangements} = \frac{13!}{2! \times 3! \times 2! \times 2!} \] \[ \displaystyle = \frac{6,227,020,800}{2 \times 6 \times 2 \times 2} = \frac{6,227,020,800}{48} = 129,729,600 \]

First, find \( \displaystyle M^T \) (rows become columns): \[ \displaystyle M^T = \begin{pmatrix} -2 & 0 & 5 \\ 0 & -1 & 6 \\ 4 & 6 & 3 \end{pmatrix} \] Next, find \( \displaystyle 2M \): \[ \displaystyle 2M = \begin{pmatrix} -4 & 0 & 8 \\ 0 & -2 & 12 \\ 10 & 12 & 6 \end{pmatrix} \] Add them together: \[\displaystyle \begin{pmatrix} -2-4 & 0+0 & 5+8 \\ 0+0 & -1-2 & 6+12 \\ 4+10 & 6+12 & 3+6 \end{pmatrix} = \begin{pmatrix} -6 & 0 & 13 \\ 0 & -3 & 18 \\ 14 & 18 & 9 \end{pmatrix} \]

First, calculate \( \displaystyle \sum f \) and \( \displaystyle \sum fx \): \[ \displaystyle \sum f = 5+7+3+7+11+6+7 = 46 \] \[ \displaystyle \sum fx = (0\times 5) + (1\times 7) + (2\times 3) + (3\times 7) + (4\times 11) + (5\times 6) + (6\times 7) \] \[ \displaystyle \sum fx = 0 + 7 + 6 + 21 + 44 + 30 + 42 = 150 \] \[ \displaystyle \text{Mean } (\bar{x}) = \frac{150}{46} \approx 3.26 \]

Variance \( \displaystyle \sigma^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 \). \[ \displaystyle \sum fx^2 = (0^2\cdot 5) + (1^2\cdot 7) + (2^2\cdot 3) + (3^2\cdot 7) + (4^2\cdot 11) + (5^2\cdot 6) + (6^2\cdot 7) \] \[ \displaystyle = 0 + 7 + 12 + 63 + 176 + 150 + 252 = 660 \] \[ \displaystyle \sigma^2 = \frac{660}{46} - (3.26)^2 = 14.347 - 10.627 = 3.72 \]

By geometric definition, the set of all points that are at an equal distance from two intersecting lines forms the bisector of the angles formed by those lines. Any point on the angle bisector will have equal perpendicular distances to both lines.

The sum of two opposite angles of a cyclic quadrilateral = 180°

\(\therefore\) (2x + 18)° + 84° = 180°

2x + 102° = 180° \(\implies\) 2x = 78°

x = 39°

Solve the equation for \( \displaystyle \sin x \): \[ \displaystyle 4\sin^2 x = 3 \] \[ \displaystyle \sin^2 x = \frac{3}{4} \] \[ \displaystyle \sin x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] For the range \( \displaystyle 0^\circ \le x \le 90^\circ \), the angle whose sine is \( \displaystyle \frac{\sqrt{3}}{2} \) is \( \displaystyle 60^\circ \).

Area of rectangle ABCD = length x breadth

= 7 x 4 

= 28 cm\(^2\)

Area of triangle CDE = \(\frac{1}{2}\) base x height

= \(\frac{1}{2} \times 3 \times 4\)

= 6 cm\(^2\)

Area of the figure = 28 cm\(^2\) + 6 cm\(^2\)

= 34 cm\(^2\)

To find the total marks, we multiply each score (x) by the number of students (f) who got that score and then sum them up (\( \displaystyle \sum fx \)): \[ \displaystyle (0 \times 4) + (1 \times 3) + (2 \times 7) + (3 \times 8) + (4 \times 6) + (5 \times 2) \] \[ \displaystyle = 0 + 3 + 14 + 24 + 24 + 10 = 75 \]