\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
Control limits for mean chart
UCL = \(\bar{\bar{X}}\) +  A_{2^\(\bar{R}\)}
= 10.66 + .577(6.3) = 14.295
CL = \(\bar{\bar{X}}\) = 10.66
Control limits for R-chart
UCL =   D_{2^\(\bar{R}\)} = 2.115 \(\times\) 6.3
= 13.324
CL = _^\(\bar{R}\) = 6.3
LCL = D₃\(\bar{R}\) = 0

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 n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

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 = \(\frac{1}{15}.\)

Max (9, 8, 5) = 9
∴ d₁ is the best strategy using maximin principle.

\(\int { x } \sqrt { x+2 } dx\) = \(\int { (x+2-2) } \sqrt { x+2 } dx\)
[Adding & subtracting 2]
= \(\int { (x+2) } \sqrt { x+2 } dx-\int { 2 } \sqrt { x+2 } dx\)
= \(\int { { \left( x+2 \right) }^{ \frac { 3 }{ 2 } } } dx-2\int { { \left( x+2 \right) }^{ \frac { 3 }{ 2 } } } dx\)
= \(\frac { { \left( x+2 \right) }^{ \frac { 3 }{ 2 } +1 } }{ \frac { 3 }{ 2 } +1 } -2\frac { { \left( x+2 \right) }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +c\)
= \(\frac { { \left( x+2 \right) }^{ \frac { 5 }{ 2 } } }{ \frac { 5 }{ 2 } } -2\frac { { \left( x+2 \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c\)
= \(\frac { 2 }{ 5 } { \left( x+2 \right) }^{ \frac { 5 }{ 2 } }-\frac { 4 }{ 3 } { \left( x+2 \right) }^{ \frac { 3 }{ 2 } }+c\)

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

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 ηd = 1
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =1\)
\(\Rightarrow \frac { dx }{ x } =\frac { -dp }{ p } \)
⇒ log x = log p + log k
⇒log x+ log p = log k
⇒ log px = log k
⇒ px = k
Demand function P = \(\frac{k}{x}\)

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right| =12-12=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\therefore \Delta =\Delta x=\Delta y=0\)
\(\therefore \) The system is consistent with infinite number of solutions
let y = k, \(k\epsilon R\)
\(\therefore 2x+3k=4\Rightarrow 2x=4-3k\)
\(\Rightarrow x=\cfrac { 1 }{ 2 } \left( 4-3k \right) ,k\epsilon R\)
\(\therefore \) Solution set is \(\left\{ \cfrac { 4-3k }{ 2 } ,k \right\} ,k\epsilon R\)