₹28,000

(i) This requires a complete list of the population but such upto date lists are not available in many enquiries.
(ii) If the size of the sample is small, then it will not be a representative of the population.

Let X be the random variable denoting the number of defective items
\( \mathrm{p}=\frac{3}{100}, \mathrm{n}=100, \lambda=\mathrm{np}=3 \)
\(\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-\dot{\lambda}} \lambda^{x}}{x !} \)
\(\mathrm{P}(\mathrm{X}=0)=\frac{e^{-3} 3^{0}}{0 !}=0.0498 \)
\(\mathrm{P}(\mathrm{X}=1)=\frac{e^{-3} 3^{1}}{1 !}=3(0.0498)=0.1494 \)

\(\mathrm{F}(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\)
\( P(0 \leq x \leq 1) =\mathrm{F}(1)-\mathrm{F}(0) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right) \)
\( =\frac{1}{2}+\frac{1}{\pi}\left(\frac{\pi}{4}\right)-\frac{1}{2}-\frac{1}{\pi}(0)=\frac{1}{4} \)

\( \mathrm{A} =\int_{a}^{b} y d x \quad \frac{x=0 \mid}{(0,0)} \mid(2,0) \)
\(=\int_{0}^{2}\left(4 x^{2}-8 x+6\right) d x \)
\( =\left[\frac{4 x^{3}}{3}-\frac{8 x^{2}}{2}+6 x\right]_{0}^{2}=\frac{32}{3}-16+12 \)
\( =\frac{32}{3}-4=\frac{20}{3} \text { sq. units } \)

\(\int \frac{e^{5 \log _{e} x}-e^{4 \log _{e} x}}{e^{3 \log _{e} x}-e^{2 \log _{e} x}} d x=\int \frac{e^{\log _{e} x^{3}}-e^{\log _{e} x^{4}}}{e^{\log _{e} x^{3}}-e^{\log _{e} x^{2}}} d x\)
\( =\int \frac{x^{5}-x^{4}}{x^{3}-x^{2}} d x=\int \frac{x^{2}\left(x^{3}-x^{2}\right)}{x^{3}-x^{2}} d x \\ \)
\(=\int x^{2} d x=\frac{x^{3}}{3}+c \)

\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
∴ order = 3, 
∴ Degree = 1

Max (6, 2, 4) = 6
∴ Action Al is the best according to maximin principle

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
 We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)

Let \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
 Consider the second order minor
\(\left| \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right| =-6-(-3)\)
= -6 + 3 = -3
\(\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

LHS = E∇

[∵ Δ = E - 1]
= E - 1 = Δ
RHS

= E - 1 = Δ = RHS
∴ E∇ = Δ = ∇ E
Hence proved.

\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\Delta \left( \Delta \left( \frac { 1 }{ x } \right) \right) \)
Now \(\Delta \left[ \frac { 1 }{ x } \right] =\frac { 1 }{ 1+x } -\frac { 1 }{ x } \)
\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) ={ \Delta }\left( \frac { 1 }{ 1+x } -\frac { 1 }{ x } \right) \)
\(=\Delta \left( \frac { 1 }{ 1+x } \right) -\Delta \left( \frac { 1 }{ x } \right) \)
Similarly \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\frac { 2 }{ x(x+1)(x+2) } \)

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.

Given  \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D²−4D+5)y = 0
The auxiliary equation is m²−4m + 5 = 0
⇒ (m−2)²−4 + 5 = 0
(m− 2)² = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e^2x[A cos x + B sin x]
The general solution is y = e^2x[A cos x + B sin x]

Cyclic uniformly periodic in nature. They may or may not follow exactly similar patterns after equal intervals of time. Generally one cyclic period ranges from 7 to.9 years and there is no hard and fast rule in the fixation of years for a cyclic period. For example, every business cycle has a Start-Boom-Depression- Recover maintenance during booms and depressions, changes in government monetary policies, changes in interestrates.

Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and it is usually denoted by H₁.

The statistical constants of the population like mean (μ), variance (σ²) are referred as population parameters.

Normal distribution is a limiting case of Binomial distribution under the following conditions.
(i) n, the number of trials is infinitely large i.e. n ⟶ ∞
(ii) Neither p(or q) is very small.

Poisson distribution is a limiting case of binomial distribution under the following conditions.
(i) n, the number of trials is indefinitely large ie., n⟶∞.
(ii) p, the constant probability of success in each trial is very small ie., p ⟶0.
(iii) np = λ is finite.Thus p = λ/n and q = 1 -(λ/n) where λ is a positive real number.

When we perform any experiment, we expect an outcome. We associate a real numbers with each outcome of an experiment. In other words, we considering a function whose domain is the set of possible outcomes and whose range is subset of the set of real numbers such a function is called random variable.

The given probability function is

Since the given function is a probability function, each ρ_i>0 and Σρ_i = 1
⇒ k + 2k + 3k + 4k = 1
⇒10k = 1 ⇒ k = \(\frac{1}{10}\)
⇒ k = 0.1

2y + x = 8

Given 2y + x = 8
2y = 8-x
y = \(\frac{1}{2}\) (8-x)
Given limits are x = 2 and x = 4
Area of the shaded region between the given limits
\(A=\int _{ a }^{ b }{ y\quad dx } =\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } (8-x)dx } \)
\(\frac { 1 }{ 2 } \int _{ 2 }^{ 4 }{ (8-x)dx } =\frac { 1 }{ 2 } { \left[ 8x-\frac { { x }^{ 2 } }{ x } \right] }_{ 2 }^{ 4 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 8(4)-\frac { { 4 }^{ 2 } }{ 2 } \right) \left( 8(2)-\frac { { 2 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } [(32-8)-(16-21)]\)
\(=\frac { 1 }{ 2 } [24-14]\)

A = 5 sq. units.

\(\int { \sqrt { 1-\sin2x } dx } \)
\(=\int { \sqrt { { \sin }^{ 2 }x+{ \cos }^{ 2 }x-2\sin x \cos x } dx } \) [∵1 = sin²x + cos²x sin²x = 2sin x cos x]
\(=\int { \sqrt { { \left( \sin x-\cos x \right) }^{ 2 } } } dx\) [∵(a−b)²=a²−2ab+b²]
=∫(sinx − cosx)dx
=−cosx − sinx+ c
=−(cosx + sinx)+c

\(\int { \left( { x }^{ 3 }+7 \right) \left( x-4 \right) dx } \)
= \(\int { \left( { x }^{ 4 }-{ 4x }^{ 3 }+7x-28 \right) dx } \)
\(= \frac { { x }^{ 5 } }{ 5 } -{ x }^{ 4 }+\frac { { 7x }^{ 2 } }{ 2 } -28x+c\)

Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)