The table below shows marks obtained by a form three class in a certain school. Marks (x): 2, 6, 8, 3, 1 No. of students (y): 2, 6, 8, 3, 1 Use the table to represent the information on a histogram. (

The horizontal axis (x-axis) represents the 'Marks (x)', and the vertical axis (y-axis) represents the 'Number of students (y)' (frequency).

For the marks, you can use intervals of 1 unit. For example, mark 1 can be represented by a bar from 0.5 to 1.5, mark 2 from 1.5 to 2.5, and so on.

The width of each bar should be uniform (e.g., 1 unit). The height of each bar should correspond to the frequency (number of students) for that mark.

• For Mark 1, draw a bar of height 1.
• For Mark 2, draw a bar of height 2.
• For Mark 3, draw a bar of height 3.
• For Mark 6, draw a bar of height 6.
• For Mark 8, draw a bar of height 8.

There will be gaps where there are no marks (e.g., between mark 3 and mark 6, and between mark 6 and mark 8).

16. Find the inverse of the matrix and hence solve the simultaneous equations below. (4mks)
$x + y = 21$
$2x + y = 21$

Let $A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, and $B = \begin{pmatrix} 21 \\ 21 \end{pmatrix}$.

The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.

The inverse $A^{-1}$ is given by $\frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

The inverse of the matrix is $\boxed{\begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix}}$.

To solve for $X$, we use the formula $X = A^{-1}B$.

Perform the matrix multiplication:

The solution to the simultaneous equations is $x = \boxed{0}$ and $y = \boxed{21}$.