leap year has 366 days. So it has 52 full weeks and 2 days. 52 Saturdays must be in 52 full weeks.
The possible chances for the remaining two days will be the sample space.
S = {(Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}
n(S) = 7
Let A be the event of getting 53^rd Saturday.
Then A = {Fri-Sat, Sat-Sun}; n(A) = 2
Probability of getting 53 Saturdays in a leap year is P(A0 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

Standard deviation \(\sigma=6.5\)
Mean \(\bar{x}=12.5\)
Coefficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \% \)
\(=\frac{6.5}{12.5} \times 100 \% \)
\(=\frac{65}{125} \times 100 \% \)
\(=\frac{13}{25} \times 100 \% \)
= 52 %
Co-efficient of variation is 52%

From the right \(\triangle\)ABC
\( \tan \theta =\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{A C}{B C} \)
\(=\frac{10 \sqrt{3} m}{30 m}=\frac{\sqrt{3}}{3} \)
\(=\frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\tan \theta =\frac{1}{\sqrt{3}} \)
\(\theta =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
0 = 30^o
Angle of elevation is 30^o

If the range = 36.8 and
the smallest value =13.4
Range R = L - S
36.8 = L-13.4
= 36.8 + 13.4 = 50.2
The largest value L = 50.2

\(\sqrt { 256{ \left( x-a \right) }^{ 8 }{ \left( x-b \right) }^{ 4 }{ \left( x-c \right) }^{ 16 }{ \left( x-d \right) }^{ 20 } } \) = 16|(x - a)⁴(x - b)² (x - c)⁸ (x - d)¹⁰|

\(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } = \frac { \frac { 1 }{ cos\theta } }{ sin\theta } -\frac { sin\theta }{ cos\theta } =\frac { 1 }{ sin\theta cos\theta } -\frac { sin\theta }{ cos\theta } \)
\(=\frac { 1-si{ n }^{ 2 }\theta }{ sin\theta cos\theta } =cot\theta \)

36x⁴ - 60x³ + 61x² - mx + n
= (6x²)² - 2.6x².5x + (5x)² + 36x² - mx + n
= (6x² - 5x)² + 36x² - mx + n
= (6x² - 5x)² + 2.3.(6x² - 5x) + 30x - mx + n
= (6x² - 5x)² + 2.3.(6x² - 5x) + 3² + (30 - m)x
= (6x² - 5x + 3)² + (30 - m)x + (n - 9)
Therefore the last obtained polynomial will be a perfect square if
30 - m = 0 and n - 9 = 0
Now
30 - m = 0 gives m = 30
n - 9 = 0 gives n = 9

Let BC is the tower.
CD is the water tank.
In right triangle \(\Delta\)ABD
\(\tan 45^{\circ} =\frac{B D}{A B} \)
\(1 =\frac{B C+C D}{40} \)
\(B C+C D =40 \mathrm{~m} \)
\(\text { In the right triangle } \triangle A B C\)
\(\tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}} =\frac{B C}{40} \)
\(\mathrm{BC}=\frac{40}{\sqrt{3}} m \)
\(\mathrm{BC}=23.1 \mathrm{~m} \)
\(\text { Substituting } \mathrm{BC}=23.1 \mathrm{~m} \text { in (1) }\)
\(23.1+C D =40 \)
\(C D =40-23.1 \)
\(=16.9 \mathrm{~m} \)
Height of the tower = 23.1 m.
Depth of the tank = 16.9 m.

When two unbiased dice are rolled, the Sample Space
s = {(1, 1) (1, 2) (r,3) (1,4) (1,5) (1,6)
(2, 1) (2,2) (2, 3) (2, 4) (2,5) (6, 6)
(3, 1) (3,2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4,2) (4,3) (4,4) (4, 5) (4,6)
(5, 1) (5,2) (5,3) (5,4) (5,5) (6,6)
(6, 1) (6,2) (6, 3) (6,4) (6, 5) (6, 6)}
n(S) = 36
(i) Let A be the event of getting a doublet
A = {( 1, 1) (2,2) (3,3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting the product as a prime number.
B = {(1,2) (1,3) (1, 5) (2,1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting the sum as a prime number.
c = {(1, 1) ( 1, 2) (1, 4) ( 1, 6) (2, 1) (2, 3) (2, 5) (3,2) (3, 4) (4, 1) (4,3) (5,2) (5,6) (6, 1) (6,5)}
n(C) = 15
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting the sum as 1. Since it is an impossible event.
n(D) = 0 and P(D) = g

The height of the tower AB = 50 m
Let the height of the tree CD = y and BD = x
From the diagram,\(\angle \)XAC = 30° = \(\angle \)ACM and\(\angle \) = XAD = 45° =\(\angle \) ADB
In right triangle ABD,
tan45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 50 }{ x } \) gives x = 50 m
In right triangle AMC,
tan30° = \(\frac { AM }{ CM } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { AM }{ 50 } \)[since DB = CM]
AM = \(\frac { 50 }{ \sqrt { 3 } } =\frac { 50\sqrt { 3 } }{ 3 } =\frac { 50\times 1.732 }{ 3 } \) = 28.87 m.
Therefore, height of the tree = CD = MB = AB − AM = 50 – 28.87 = 21.13 m

Let the assumed mean, A = 8

Standard deviation
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 240 }{ 29 } -\left( \frac { 4 }{ 29 } \right) ^{ 2 } } =\sqrt { \frac { 240\times 29-16 }{ 29\times 29 } } \)
σ = \(\sqrt { \frac { 6944 }{ 29\times 29 } } \); σ ≃ 2.87
Calculation of Standard deviation for continuous frequency distribution
(i) Mean method:
Standard deviation σ = \(\sqrt { \frac { \Sigma { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 } }{ N } } \)
Where, x_i = Middle value of the i th class
f_i = Frequency of the i th class
(ii) Shortcut method (or) Step deviation method:
To make the calculation simple, we provide the following formula. Let A be the assumed mean, x_i be the middle value of the i^th class and c is the width of the class interval.
Let d_i = \(\frac { { x }_{ i }-A }{ c } \)
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \).

Mean \(\bar { x } \) = \(\frac { Sum\ of\ all\ observations }{ Number\ of\ observation } \)
= \(\frac { \Sigma x_{ i } }{ n } =\frac { 1+2+3+...+n }{ n } =\frac { n(n+1) }{ 2\times n } \)
Mean \(\bar { x } \) = \(\frac { n+1 }{ 2 } \)
Variance σ² = \(\frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { \Sigma x_{ i } }{ n } \right) ^{ 2 }\left[ \begin{matrix} \Sigma x_{ i }^{ 2 }={ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 } \\ (\Sigma x_{ i })^{ 2 }=(1+2+3+...+n)2 \end{matrix} \right] \)
= \(\frac { n(n+1)(2n+1) }{ 6\times n } -\left[ \frac { n(n+1) }{ 2\times n } \right] ^{ 2 }\)
= \(\frac { 2n^{ 2 }+3n+1 }{ 6 } -\frac { { n }^{ 2 }+2n+1 }{ 4 } \)
Variance σ² = \(\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { { n }^{ 2 }-1 }{ 12 } \).

Let AB be the height of the kite above the ground. Then, AB = 75.
Let AC be the length of the string.
In right triangle ABC,\(\angle \)ACB = \(60°\)
\(sin\theta =\frac { AB }{ AC } \)
\(sin60°=\frac { 75 }{ AC } \)
gives \(\frac { \sqrt { 3 } }{ 2 } =\frac { 75 }{ AC } \) so, AC = \(\frac { 150 }{ \sqrt { 3 } } =50\sqrt { 3 } \) 
Hence, the length of the string is 50\(\sqrt { 3 } m\)

Given
\( \frac{\cos \alpha}{\cos \beta}=m \)
\(\frac{\cos \alpha}{\sin \beta} =n \)
\(\text { LHS } =\left(m^{2}+n^{2}\right) \cos ^{2} \beta \)
\(=\left(\frac{\cos ^{2} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}\right) \cos ^{2} \beta \)
\(=\frac{\left(\cos ^{2} \alpha \sin ^{2} \beta+\cos ^{2} \alpha \cos ^{2} \beta\right)}{\cos ^{2} \beta \sin ^{2} \beta} \cos ^{2} \beta \)
\(=\frac{\cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)}{\sin ^{2} \beta} \)
\(=\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}(1) \)
\(=\left(\frac{\cos \alpha}{\sin \beta}\right)^{2} \)
= n² = RHS

Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 10), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)

To solve the equation X² - 5x - 14 = 0, subtract X² - 5x - 14 = 0 from y = X² - 5x - 6.
 is a straight line parallel to x axis.
The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the
equation X² - 5x - 14 = 0.
∴ Solution {-2, 7}

Point of intersection (-3,5), (4, 12) solution of x² -x - 12 = 0 is -3, 4