Further Mathematics Questions and Answers

$\overrightarrow{OY} \equiv -\overrightarrow{YO}$

Also, $\overrightarrow{YO} + \overrightarrow{OX} = \overrightarrow{YX}$

$\therefore \overrightarrow{YO} = -\overrightarrow{OY} = - \begin{pmatrix} 16 \\  -11  \end{pmatrix} = \begin{pmatrix} -16 \\ 11  \end{pmatrix}$

$\overrightarrow{YX} = \begin{pmatrix}  -16 \\ 11  \end{pmatrix}  + \begin{pmatrix}  -7  \\  6   \end{pmatrix}$

= $\begin{pmatrix}  -23  \\  17  \end{pmatrix}$

$\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}$

= $\frac{1}{2} + \frac{1}{6} = \frac{2}{3}$.

Hint: Probability of A and B occurring should be greater than probability A or B happening.

Given $a = i - 3j; b = -2i + 5j; c = 3i - j$

$a- b + c = (1 - (-2) + 3)i + (-3 - 5 + (-1))j = 6i - 9j$

$|a - b + c| = \sqrt{6^{2} + (-9)^{2}} = \sqrt{36 + 81} = \sqrt{117}$

$= \sqrt{9 \times 13} = 3\sqrt{13}$

$\rho = 1 - \frac{6\sum{d^{2}}}{n(n^{2} - 1)}$

$1 - \frac{6 \times 4}{4(4^{2} - 1)} = 1 - \frac{24}{60}$

= $1 - 0.4 = 0.6$

Let the number of mixed schools be S

The number of boys only = 42 - S

The number of girls only = 35 - S

(42 - S) + S + (35 - S) = 77 - S = 70

$\implies$ S = 7

$Probability = \frac{7}{70} = \frac{1}{10}$