Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

Here number of rows and columns are equal
∴ The given assignment problem is balanced.
Step 1 : Select a minimum element in each row and subtract this from all the elements in its row.                  

A	0	8	14	17
B	0	15	6	10
C	1	3	0	11

Here column V has no zero. Go to Step 2.
Step 2 : Select the minimum element in each column and subtract this from all the elements in its column.                          

	U	V	W
A	0	5	14
B	0	12	6
C	1	0	0
Column minimum	3

New, each row and column contains atleast one zero. Hence, assignments can be made.
Step 3 : Examine the rows with exactly one zero. Mark it by Ԡ and draw a vertical line.
After examining the row, examine the column with one zero mark it by Ԡ and draw a horizontal line.

New the elements not lying on the line are

5	14
12	6

and min is 5 subtract 5 from all these number and add 5 to 1which lies in the intersecting lines. Other numbers remains the same.
A new cost matrix will be formed and repeat step 3.

Thus, all the 3 assignments have been made.
∴The optimal assignment schedule and total cost is

Job	Machine	Cost
A	V	25
B	U	10
C	W	11

Total cost = Rs. 46

yx² dx = -e^-x dy
⇒ \(\frac { { x }^{ 2 } }{ { e }^{ -x } } dx=-\frac { dy }{ y } \)
⇒ x² e^x dx =\(-\frac { dy }{ y } \)
Integrating both sides,
\(\int { { x }^{ 2 }{ e }^{ x } } dx=-\int { \frac { dy }{ y } } \)
put u = x²
u^'= 2x
u^'' = 2
dv = e^x dx
v = e^x
v₁ = e^x
v₂ = e^x
Using Bernoulli's formula,
\(\int { u } dv\) = uv-u^'v₁ + u^''v₂
∴ \(\int { { x }^{ 2 }{ e }^{ x } } \)dx = x²e^x-2xe^x+2e^x
∴ From (1), x²e^x-2xe^x + 2e^x = -log y+C
⇒ e^x(x²-2x+2)+log y = C

Expenditure	Weights (V)	Rent 2010 (p0)	Price 2011 (p1)	P =  \(\frac {p_{1}}{p_{0}} \times 100\)	PV
Food	35	150	174	116	4060
Rent	15	50	60	120	1800
Clothing	20	100	125	125	2500
Fuel	10	20	25	125	1250
Rice	20	60	90	150	3000
	100				12610

Cost of living index number = \(\frac {\sum PV}{\sum V}\) = \(\frac {12610}{100}\) = 126.10
∴  The cost ofliving has increased upto 26.10% in 2011 as compared to 2010.

Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.
Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

S.No	Statistic	Standard Error
1	Sample mean	σ/√n
2	Observed sample proportion	\(\sqrt { PQ/n } \)
3	Sample standard deviation	\(\sqrt { { \sigma }^{ 2 }/2n } \)
4	Sample variance	\({ \sigma }^{ 2 }\sqrt { 2/n } \)
5	Sample quartiles	\(1.36263\sigma /\sqrt { n } \)
6	Sample median	\(1.25331\sigma /\sqrt { n } \)
7	Sample correlation coefficient	\((1-{ \rho }^{ 2 })/\sqrt { n } \)

Let p be the probability of a patient to recover
Given q = 75% = \(\frac { 75 }{ 100 } \)  = 0.75
∴ P = 1 - q = 1 - 0.75 = 0.25
n = 6
P (4 will recover) = P(X = 4)
= 6C₄ (0.25)⁴ (0.75)²
[∵ P(x) = nC_x p^xq^n-x n = 6, x = 4
= 6C₂ (0.25)⁴ (0.75)²

= 15 (0.25)⁴ (0.75)²
P(X = 4) = 0.03295

1) 0≤F(x)≤1, -∞
2) F(-∞) = 0 and F(∞) = 1
3) is a non-decreasing function, F(a) ≤F(b) for a
4) \(\underset { h\rightarrow 0 }{ lim } \) F(x+h) = F(x), since F(x) is continuous from the right
5) F'(x) = f(x)≥0
6) P(a≤x≤b) = F(b)-F(a)

\(\int { \frac { 1 }{ { \sin }^{ 2 }x{ \cos }^{ 2 }x } } dx\)
\(=\int { \frac { \left( { \sin }^{ 2 }x+{ \cos }^{ 2 }x \right) }{ { \sin }^{ 2 }x\ { \cos }^{ 2 }x } } dx\) [∵1 = sin²x + cos²x]

\(=\int { \frac { 1 }{ { \cos }^{ 2 }x } } dx+\int { \frac { dx }{ { \sin }^{ 2 }x } } \)
= ∫sec²x dx + ∫cosec²x dx
= tan x − cot x + c

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =10-9=1\neq 0\)
Since \(\Delta \neq 0\) 
we can apply Cramer's rule and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =7(5)-9(3)\)
= 35 - 27 = 8
\(\Delta y=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =2(9)-3(7)\)
= 18 - 21 = -3
 \(\therefore\) \(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 8 }{ 1 } =8\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
\(\therefore\) Solution set is (8, -3)

Given P_d = 18− 2x − x² ; P_s = 2x − 3
We know that at equilibrium prices p_d = p_s
18− 2x − x² = 2x – 3
x² + 4x −21 = 0
(x − 3) (x + 7) = 0
x = –7 or 3
The value of x cannot be negative, x = 3
When x₀ = 3
ஃ p₀ = 18 − 2(3)  − (3)² = 3
CS = \(\int _{ 0 }^{ { x }_{ o } }{ f(x) } \) dx - x₀p₀
= \(\int _{ 0 }^{ 3 }{ (18-2x-{ x }^{ 2 }) } \)dx - 3 x 3
= \({ \left[ 18x-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }\)- 9
= 18(3) - (3)² - \(\left( \frac { { 3 }^{ 3 } }{ 3 } \right) \) - 9
CS = 27 units
PS = x₀P₀ - \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \)
= (3 \(\times\) 3) - \(\int _{ 0 }^{ 3 }{ (2x-3) } \)
= 9 - \(({ { { x }^{ 2 }-3x) } }_{ 0 }^{ 3 }\)
= 9 units
Hence at equilibrium price,
(i) the consumer’s surplus is 27 units
(ii) the producer’s surplus is 9 units.