What is the class width of the distribution? Calculate the mean marks of the distribution.

The given class intervals are $41-49$ and $50-58$.

The class width can be found by subtracting the lower limit of one class from the lower limit of the next class, or by using the formula (upper limit - lower limit) + 1 for a single class.
Using the lower limits: $50 - 41 = 9$.
Using the first class: $(49 - 41) + 1 = 8 + 1 = 9$.
The class width is $\boxed{\text{9}}$.

Part c) Calculate the mean marks of the distribution.

To calculate the mean of a frequency distribution, we need the class mark ($x$) and frequency ($f$) for all classes in the distribution. The formula for the mean is $\bar{x} = \frac{\sum fx}{\sum f}$.

From the provided image, only the data for the first class interval ($41-49$) is visible:
Class Mark ($x$) = 45
Frequency ($f$) = 10
$fx$ = 450

The frequencies for the other class intervals (e.g., $50-58$, and $86-94$) are missing, as are the total sums $\sum f$ and $\sum fx$. Therefore, it is not possible to calculate the mean marks of the entire distribution with the information provided.

Question 19 A) Inequalities

The given inequalities are $x > 0$, $y > 0$, and $x + 2y \le 6$.
The boundary lines are:

• $x = 0$ (the y-axis)
• $y = 0$ (the x-axis)
• $x + 2y = 6$

• If $x = 0$, then $2y = 6 \implies y = 3$. Point: $(0, 3)$.
• If $y = 0$, then $x = 6$. Point: $(6, 0)$.

The region satisfying $x > 0$ is to the right of the y-axis.
The region satisfying $y > 0$ is above the x-axis.
The region satisfying $x + 2y \le 6$ is below or on the line $x + 2y = 6$.
Combining these, the region is a triangle in the first quadrant, bounded by the x-axis, the y-axis, and the line $x + 2y = 6$. The vertices of this triangular region are $(0,0)$, $(6,0)$, and $(0,3)$. The region includes the line $x+2y=6$ but not the axes themselves (due to strict inequalities $x>0, y>0$).

We need integer values for $x$ and $y$ such that $x \ge 1$, $y \ge 1$, and $x + 2y \le 6$.

• If $y = 1$: $x + 2(1) \le 6 \implies x + 2 \le 6 \implies x \le 4$. Since $x \ge 1$, the possible integer values for $x$ are $1, 2, 3, 4$. Points: $(1,1), (2,1), (3,1), (4,1)$.
• If $y = 2$: $x + 2(2) \le 6 \implies x + 4 \le 6 \implies x \le 2$. Since $x \ge 1$, the possible integer values for $x$ are $1, 2$. Points: $(1,2), (2,2)$.
• If $y = 3$: $x + 2(3) \le 6 \implies x + 6 \le 6 \implies x \le 0$. Since $x \ge 1$, there are no integer values for $x$ when $y=3$.

The integral values of $x$ and $y$ that satisfy the inequalities are:
$\boxed{(1,1), (2,1), (3,1), (4,1), (1,2), (2,2)}$

Question 19 B) Rectangle perimeter and diagonal

Let the length of the rectangle be $L$ and the width be $W$.
The perimeter $P = 2(L + W) = 46$ cm.
The diagonal $D = 17$ cm. By the Pythagorean theorem, $L^2 + W^2 = D^2$.

From the perimeter:

From the diagonal:

From equation (1), express $L$ in terms of $W$:

Substitute this into equation (2):

Divide the entire equation by 2:

We can factor the quadratic equation. We need two numbers that multiply to 120 and add up to $-23$. These numbers are $-8$ and $-15$.

This gives two possible values for $W$:

• If $W = 8$ cm, then $L = 23 - 8 = 15$ cm.
• If $W = 15$ cm, then $L = 23 - 15 = 8$ cm.

Conventionally, the width is considered the shorter side of the rectangle.
Therefore, the width of the rectangle is $\boxed{\text{8 cm}}$.