Arranging in ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
n = 11
 Q₁ = size of \((\frac { { n+1} }{ 4 } )^{ th }\)  value
= size of \({ \left( \frac { 11+1 }{ 4 } \right) }^{ th }\) value
= size of 3^rd value = 6
Q₃ = Size of \({ \left( \frac { 3(n+1) }{ 4 } \right) }^{ th }\) value
= size of \({ \left( 3\left( \frac { 11+1 }{ 4 } \right) \right) }^{ th }\) value
=size of 3(3)^th value
= size of 9^th value = 18

Here a = 18,000 ; i = 0.15 and k = 12.
P = \({\frac{a}{{i}/k}} = {\frac{18,000}{0.15/12}}\) 
= \({\frac{18,000}{0.15}} \times {12}\)
\(=\frac{18,00,000\times 12}{15}=14,40,000\)
Therefore the amount to be deposited is Rs. 14,40,000.

Given n = 15, \(\sigma_x=3.01, \sigma_y=3.03 \)
\(\Sigma(X-\bar{X})(Y-\bar{Y})=122 \)
\(r(x, y) =\frac{1}{N} \frac{\Sigma(X-\bar{X})(Y-\bar{Y})}{\sigma_x \sigma_y} \)
\(=\frac{122}{15 \times 3.01 \times 3.03} \)
\(=\frac{122}{136.8045}=0.8918\)

The required network for the above information.

xpⁿ = k ⇒ x = kp^-n
\({dx\over dp}=-nkp^{-n-1}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{p\over kp^{-n}}(-nkp^{-n-1})\) = n, which is a constant.

Equation of a circle when end points of the diameter are given is
(x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0
Here (x₁, y₁) = (4,7) and (x₂, y₂)  = (-2,5) 
(x - 4) (x + 2) + (y - 7) (y - 5 ) = 0
x² + 2x - 4x -8 + y² - 5y - 7y + 35 = 0
x² + y² - 2x - 12y + 27 = 0

The letter of the word CHAT in alphabetical order are A, C, H, T.
(i) Number of words starting with A = 3! = 6 C begins
(ii) Number of words starting with CA are 2! = 2
Now the words CH begins
(iii) After that we get the word CHAT = 1
\(\therefore\) Rank of CHAT is = 6 + 2 + 1 = 9

\({ -320 }^{ o }=-320\times \frac { \pi }{ 180 } \)= \(\frac { -16\pi }{ 9 } \)

\(f(x)=log({ x }^{ 2 }+\sqrt { { x }^{ 2 } } +1)\)
\(f(-x)=log(-{ x }^{ 2 }+\sqrt { { (-x) }^{ 2 }+1 } )\)
= log  \(({ x }^{ 2 }+\sqrt { { x }^{ 2 }+1 } )\) = f(x)
\(\therefore\) f is an even function

The technology matrix is
\(B=\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}\)
I - B =\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.25 \\ -0.40 & 0.33 \end{bmatrix}\)
\(|\mathrm{I}-\mathrm{B}|\) = 0.165 - 0.1 = 0.065 > 0
Since the diagonal elements of (I - B) are positive and |I - B| is positive, Hawkins - Simon conditions are satisfied.
Therefore the given system is viable.