Cost of living index (C.L.I) = \(\frac{\Sigma PV}{\Sigma V}\)
= \(\frac{15800}{100}\) = 158
Hence, there is 58% increase in cost of living in 2003 compared to 2000.

Given sample size n = 400
Sample mean \(\bar { x } \) = 171.38
Population mean μ = 171.17
Population Standard deviation σ = 3.3
Null hypotheses: H₀ : μ = 171.17
Alternative hypotheses: H₁ : μ ≠ 171.17
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 171.38-171.17 }{ \frac { 3.3 }{ \sqrt { 400 } } } \)
=\(\frac { 0.21 }{ 0.165 } \) = 1.273
As the level of significance is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z < \(Z_{ \frac { \alpha }{ 2 } }\) as 1.273 < 1.96
Inference: since z < \(Z_{ \frac { \alpha }{ 2 } }\), we accept the null hypotheses at 5% level of significance.
Hence, we can conclude that the sample of 400 has taken from the population with mean height of 171.17 cm.

Given mean 20 ⇒ np = 20
S.D = 4 ⇒ \(\sqrt { npq } \) = 4
∴ \(\frac { npq }{ np } =\frac { 16 }{ 20 } \Rightarrow q=\frac { 4 }{ 5 } \)
P = 1-q =\(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substitutingp and q in npq = 16, we get
\(n\times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } \) =16
n = \(\frac { 16\times 5\times 5 }{ 4 } \)  = 100
∴ Number of events = 100

Since f(x) is a probability density function,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ \infty }{ { ce }^{ -ax }dx=1 } \Rightarrow C\int _{ 0 }^{ \infty }{ { e }^{ -ax }dx=1 } \)
\(\Rightarrow c{ \left[ \frac { { e }^{ -ax } }{ -a } \right] }_{ 0 }^{ \infty }=1\Rightarrow \frac { -c }{ a } [{ e }^{ -\infty }-{ e }^{ 0 }]\)
\(\Rightarrow \frac { -c }{ a } [0-1]=1\quad [\because { e }^{ -\infty }=0,{ e }^{ 0 }=1]\)
\(\Rightarrow \frac { c }{ a } =1\Rightarrow C=a\quad \therefore C=a\)

A bag contains 2 bad eggs and 8 good eggs
∴ Total number of eggs = 10
We are going to select 3 eggs, out of that 2 must be bad eggs.
∴ Required probability \(=\frac { { 2C }_{ 2 }\times { 8C }_{ 1 } }{ 10{ C }_{ 3 } } =\frac { 1\times 8 }{ \frac { 10\times 9\times 8 }{ 3\times 2\times 1 } } \)
\(=\frac { 1\times 8\times 3\times 2\times 1 }{ 10\times 9\times 8 } =\frac { 1 }{ 15 } \)
∴ Probability of getting two bad eggs = \(\frac{1}{15}.\)

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

Given f'(x) = 8x³ -2x²
∴ ∫ f'(x) dx = ∫ (8x³ - 2x²) dx
⇒ f(x) = \(\frac { { 8x }^{ 4 } }{ 4 } -\frac { { { 2x }^{ 3 } } }{ 3 } +c\)
⇒ f (x) = 2x⁴ - \(\frac { { { 2x }^{ 3 } } }{ 3 } +c\) ...(1)
Also, f(2) = 1
⇒ 1 = 2(2⁴) - \(\frac { { 2\left( { { 2 }^{ 3 } } \right) } }{ 3 } +c\)
⇒ 1 = 32 - \(\frac { { 16 } }{ 3 } +c\)
⇒ 1 - 32 + \(\frac { { 16 } }{ 3 } \) = c ⇒ -31 + \(\frac { { 16 } }{ 3 } \) =c
⇒ \(\frac { { -93+16 } }{ 3 } =c\)
⇒ c = \(\frac { { -77 } }{ 3 } \)
∴ (1) ⟶ f(x) = 2x⁴ - \(\frac { { 2x }^{ 3 } }{ 3 } -\frac { { -77 } }{ 3 } \)

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

Given C'(x) = 85 + \(\frac{375}{x^2}\).
We know C(x) ഽC'(x) + k
The cost of producing 10 incremental units after 15 units have been produced
\(C'(x)=85+\frac { 375 }{ { x }^{ 2 } } \)
\(C(x)=\int { C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ \left( 85+\frac { 375 }{ { x }^{ 2 } } \right) } dx\)
\({ \left[ 85+\frac { 375 }{ { x } } \right] }_{ 15 }^{ 25 }\)
\(\left( 85(25)-\frac { 375 }{ 25 } \right) -\left( 85(15)-\frac { 375 }{ 15 } \right) \)
= (2125 - 15) - (1275 - 25)
= 2110 - 1250 = Rs. 860

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions