ME

P = 0.015
n = 200
mean =\lambda =np= 200 \times 0.015 = 3
variance =\sigma ^2=\lambda =3

(a) Graph of y = 2 - x

(b) Graph of y = |2 - x|

(c) Graph of y = |2 - |x||

\begin{aligned} x_1&=45 &...(i)\\ x_2&= 50&...(ii)\\ 10x_1+10x_2&=600 \\ or\; x_1+x_2&=60&...(iii) \\ 25x_1+5x_2&=750 \\ or\; 5x_1+x_2&=150&...(iv) \\ \end{aligned}
By drawing the curve we get 3 values of x_1 and x_2 as (10, 50), (20, 50), (22.5, 37.5)
So, Z_{max}=45x_1+60x_2 for (10,50)
Z_{max}=450+3000=4450
for (20,50)
Z_{max}=45\times 20+50 \times 60=3900
for (22.5, 37.5)
Z_{max}=45\times 22.5+60 \times 37.5=3262.5
So, Z_{max}=3900\; for \; (x_1,x_2)=(20,50)

3-2-1 principle of location
The 3-2-1 principle of location (six point location principle) is used to constrain the movement of workpiece along the three axes XX, YY and ZZ.
This is achieved by providing six locating points, 3-pins in base plate, 2-pins in vertical plane and 1-pin in a plane which is perpendicular to first two planes.

Take one element {1}
T=\phi ,S=\phi \rightarrow (\phi, \phi ) \Rightarrow 1\; pair

T=1,S=\phi \rightarrow (\phi, 1 ),(1,1) \Rightarrow 2\; pair
For 1 element total pair = 3^1
Similarly
For 2 element total pair = 3^2
For 3 element total pair = 3^3
For 4 element total pair = 3^4
For 5 element total pair = 3^5
For 6 element total pair = 3^6=729

\begin{aligned} p^2-4q &\lt 4 \\ p^2-4&\ lt 4q \;\;...(i)\\ 3p+2Q& \lt 6 \\ 6p-12 &\lt -4q\;\;...(ii)\\ &\text{by equation (i) + (ii)} \\ p^2+6p-16 &\lt 0 \\ (p+8)(p-2)& \lt 0 \\ \therefore \; p&\in (-8,2) \end{aligned}
Given p is positive integer
\therefore \;p=1
Now, from equation (i), 1-4 \lt 4q
q \gt \frac{-3}{4}
from equation (ii),
q \lt 3/2
\therefore \; \frac{-3}{4} \lt q \lt \frac{3}{2}
Given q is positive integer
\therefore \;\; q=1
Thus p + q = 1+1 = 2

After 09:15:00 AM every minute, minute and second hand cross each other once (1) times.
So, 09:16:00 to 09:45:00 minute hand and second hand cross each other 30 times.

Percentage of grade III
\frac{ \text{7\% of 504 + 23 \% of 25951} }{504 + 25951}=22.7 \%

Take one element {1}
T=\phi ,S=\phi \rightarrow (\phi, \phi ) \Rightarrow 1\; pair

T=1,S=\phi \rightarrow (\phi, 1 ),(1,1) \Rightarrow 2\; pair
For 1 element total pair = 3^1
Similarly
For 2 element total pair = 3^2
For 3 element total pair = 3^3
For 4 element total pair = 3^4
For 5 element total pair = 3^5
For 6 element total pair = 3^6=729

\begin{aligned} p^2-4q &\lt 4 \\ p^2-4&\ lt 4q \;\;...(i)\\ 3p+2Q& \lt 6 \\ 6p-12 &\lt -4q\;\;...(ii)\\ &\text{by equation (i) + (ii)} \\ p^2+6p-16 &\lt 0 \\ (p+8)(p-2)& \lt 0 \\ \therefore \; p&\in (-8,2) \end{aligned}
Given p is positive integer
\therefore \;p=1
Now, from equation (i), 1-4 \lt 4q
q \gt \frac{-3}{4}
from equation (ii),
q \lt 3/2
\therefore \; \frac{-3}{4} \lt q \lt \frac{3}{2}
Given q is positive integer
\therefore \;\; q=1
Thus p + q = 1+1 = 2

Given, A P=\alpha^{2} P
By comparison with A X=\lambda X \Rightarrow
\Rightarrow \quad \lambda=\alpha^{2}
Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.

By using partial fraction concept.
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}

Mean= np
Variance = npq = np(1 - p)

On the basis of nature of carbon present in cast iron, it may be divided into white cast iron and gray cast iron.
In the gray cast iron, carbon is present in free form as graphite. Under very slow rate of cooling during solidification, carbon atoms get sufficient time to separate out in pure form as graphite. In addition, certain elements promote decomposition of cementite. Silicon and nickel are two commonly used graphitizing elements.
In white cast iron, carbon is present in the form of combined form as cementite. In normal conditions, carbon has a tendency to combine with iron to form cementite.

Particle Size, Shape, and Distribution:
Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:
1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.
2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.
3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.
4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.
5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}

\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}

\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}

\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}
It is recursion relation between y_{n+1} and y_{n}
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.