\(f(x)= \begin{cases}\frac{1}{30} & 0
Mean =\(E(X)=\int _{ 0 }^{ 30 }{ x3f(x)dx } \) 
= \(\int _{ 0 }^{ 30 }{ x.\frac { 1 }{ 30 } dx } \) 
\(E(X)=\frac { 1 }{ 30 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 30 }\) 
= \(\frac { 1 }{ 30 } [ \frac{30\times 30}{2}-0]\)  
E(X) = 15 minutes
The average waiting time for the student is 15| minutes.

We know that volume of a sphere is given by V = \(\frac43\) π r³. where r > 0 is the radius. So the differential dV = 4 ㅠr² dr and hence
Δ ≈ dv = 4π(10)² (9.9-10) cm³
= 4π10² (-0.1) cm³
= −40π cm³ 
Note that we have used dr = (9.9 −10) cm, because radius decreases from 10 to 9.9. Again the negative sign in the answer indicates that the volume of the sphere decreases about 40π cm³.

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

As \(\underset { x\rightarrow -{ 4 }^{ + } }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =-\infty \ and\ \underset { x\rightarrow { 4 }^{ + } }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\infty \) 
Therefore x = −4 and x = 4 are vertical asymptotes 
As \(\underset { x\rightarrow \infty }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\underset { x\rightarrow \infty }{ lim } \frac { 2-\frac { 8 }{ { x }^{ 2 } } }{ 1-\frac { 16 }{ { x }^{ 2 } } } =2\) and \(\underset { x\rightarrow -\infty }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\underset { x\rightarrow -\infty }{ lim } \frac { 2-\frac { 8 }{ { x }^{ 2 } } }{ 1-\frac { 16 }{ { x }^{ 2 } } } =2\)
Therefore, y = 2 is a horizontal asymptote.
This can also be obtained by synthetic division

The curve y = sin x intersects the positive x -axis. When y = 0 which gives, x = 
\( x=n\pi , n=1,2,3,...\)
Now, \(\frac{dy}{dx}=cos x\). The slpoe \(x=n\pi\)  are \(cos(n\pi)=(-1)^{n}\).   
Hence, the required angle of intersection is m₂ = 0
\(tan \theta = \frac{(-1)^n - 0}{1+((-1)^n(0)} = 1 ∀ n\)

We have,
\(\frac{d}{dt}W(t)=-20\times(1-0.1t)\)
Therefore at t = 2, \(\frac{d}{dt}W(t)=-16\)
That is, the person forgets at the rate of 16 words after 2 days of studying.

General equation of a straight line is 
ax + by + c = 0 .......(1)
where a, b, c \(\in\) R.
Since, the lines are non - vertical,we have b \(\neq\) 0
Dividing b' by equation (1),
\(( \frac{a}{b})x+y+(\frac{c}{b}) = 0 \)
\(Ax+y=C = 0, where A = \frac{a}{b}, C = \frac{c}{b}\) .....(2)
Thus, eventhough 3 arbitrary constants (a, b, c) are present in (1), they can be considered as 2 constants only, as above (2).
Differentiating (1) with respect to x
a + b \(\\ \frac { dy }{ dx } =0\)
Differentiating again with respect to 'x' we get,
(b) \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\quad [\because b\neq 0]\) .....(3)
This is the differential equation of family of all non - vertical lines in a plane.

\(-5\pi \le x\le 5\pi \) and cos x = -1
cos x = -1
\(cosx=cos\pi \)  \([\because cos\theta =cos\alpha \Rightarrow 2n\pi +\alpha ,n\varepsilon Z]\)
\(x=2n\pi +\pi ,n\varepsilon Z\quad \)
= \((2n+1)\pi ,n\varepsilon Z\)
\(\Rightarrow But-5\pi \le x\le 5\pi \)
Putting,\(=0,\pm 1,\pm 2\) and -3 we get
\(\Rightarrow x=-5\pi ,-3\pi ,-\pi ,3\pi ,5\pi \)
\(x=(2n+1)\ {\pi},n=0,\pm 1,\pm 2 \) and -3

\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 3
∴ \(\rho \)(A) ≤ min(3, 3) = 2
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right| =1\left| \begin{matrix} 4 & -6 \\ 1 & -1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} \right| \)
[Expanded along R₁]
= 1(-4+6)+2(-2+30)+3(2-20)
= 1(2)+2(28)+3(-18)
= 2+56-54 = 58-54 = 4 ≠ 0
∴ \(\rho \)(A) = 3

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.

Clearly there are 2 sign changes for the given polynomial P(x) and hence number of positive roots of P(x) cannot be more than two. Further, as P(-x) = -9x⁹- 2x⁵- x⁴- 7x²+ 2, there is one sign change for P(-x) and hence the number of negative roots cannot be more than one. Clearly 0 is not a root. So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

The distance between origin to z = i, −2 + i, and 3 are
|z| = |i| = 1
|z| = |−2+i| = \(\sqrt { \left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
|z| = |3| = 3
Since \(1<\sqrt { 5 } <3\), the farthest point from the origin is 3 .

Let cot−1\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) =\alpha \). Then, cot \(\alpha =\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \) and α is acute.
We construct a right triangle with the given data.
From the triangle, sec\(\alpha=\frac{x}{1}=x\). Thus, \(\alpha\) = sec^-1x
Hence, cot⁻¹\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

First, we find |adj (A)| = \(\left| \begin{matrix} 7 & 7 & -7 \\ -1 & 11 & 7 \\ 11 & 5 & 7 \end{matrix} \right| \) =  7(77 - 35) - 7(-7 - 77) - 7(-5 - 121) = 1764 > 0
So, we get
A = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(adj A) = \(\pm \frac { 1 }{ \sqrt { 1764 } } { \left[ \begin{matrix} +\left( 77-35 \right) & -\left( -7-77 \right) & +\left( -5-121 \right) \\ -\left( 49+35 \right) & +\left( 49+77 \right) & -\left( 35-77 \right) \\ +\left( 49+77 \right) & -\left( 49-7 \right) & +\left( 77+7 \right) \end{matrix} \right] }^{ T }\)
= \(\pm \frac { 1 }{ 42 } { \left[ \begin{matrix} 42 & 84 & -126 \\ -84 & 126 & 42 \\ 126 & -42 & 84 \end{matrix} \right] }^{ T }=\pm \left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2 \end{matrix} \right] \).

order 2;degree not defined

order – 2 : degree 2

\(\frac { 208{ a }^{ 2 } }{ 3 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { { x }^{ 3 } }{ 3 } +c\)

(i) π
(ii) πX2^n-1
(iii) 2
(iv) 0

2x – y + 3 = 0

(i) If 6 + 2 = 5, then the milk is white.
Let p: 6 + 2 = 5 (F)
q: Milk is white (T)  
p ➝ q is having the truth value T
(ii) China is in Europe or \(\sqrt3\) is an integer.
p: China is in Europe (F)
q: \(\sqrt3\) is an integer (F)
p v q is having the truth value (F).
(iii) It is not true time 5 + 5 = 9 or Earth is a planet.
Let P: 5 + 5 = 9 is not true (T)
q: Earth is a planet (T
~p ∨ q is having the truth value T
(iv) 11 is a prime number and all the sides of a rectangle are equal.
p:11 is a prime number (T)
q: Allthe sides of arectangle areequal (F)
p _^ q is having the truth value F

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

Given = (3 + sin(2x)) ^2/3 
Taking differentilas,
dy =  \(\frac23\)(3 + sin(2x)) ^2/3-1 (cos 2x) (2)dx
dy = \(\frac { 4 }{ 3 } .\frac { cos2x }{ { (3+sin2x) }^{ \frac { 1 }{ 3 } } } dx\)

If we put directly x = 1 we observe that the given function is in an indeterminate form \(\frac{0}{0}\). As the numerator and the denominator functions are polynomials of degree 2 they both are differentiable.
Hence, by an application of the l’Hôpital Rule, we get
\(\underset{x\rightarrow 1}{lim}(\frac{x^{2}-3x+2}{x^{2}-4x+3})=\underset{x\rightarrow 1}{lime}(\frac{2x-3}{2x-4})\)
= \(\frac{1}{2}\)
Note that this limit may also be evaluated through the factorization of the numerator and denominator as \(\frac{x^{2}-3x+2}{x^{2}-4x+3}=\frac{(x-1)(x-2)}{(x-1)(x-3)}\) 

Let P represent the vapour pressure and T represent the vapour temperature.
Given \(\frac { dp }{ dt } \alpha \quad p.\frac { 1 }{ { T }^{ 2 } } \)
[∴ Inversely proportional to the square of the  temperature]
\(\Rightarrow \frac { dp }{ dt } =\frac { kP }{ { T }^{ 2 } } \) where k is a constant.

Given circle is x² + y² - 8x - 9y + 12 = 0
Length of the tangent = \(\sqrt { { 2 }^{ 2 }+({ -3) }^{ 2 }-8(2)-9(-3)+12 } \)
= \(\sqrt { 4+9-16+27+12 } \)
= \(\sqrt { 36 } \) = 6 unit

We know that sin^-1:  [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
\({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 3 } \right) \right) \) = -\(\frac{\pi}{3}\), since -\(\frac{\pi}{3}\) \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

\(\sqrt { 3 } -i\)

r = 2 and \(-\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =-\alpha =-\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of
\(-\sqrt { 3 } -i\) are 2 and \(\frac { \pi }{ 6 } \)

\(\left| 2z+2-4i \right| =2\)
2|z+1-2i| = 2
⇒ |z-(-1+2i)| = 1
It is of the form |z - z₀| = r and so it represents a circle.
Its centre is (-1+2i) and radius is 1.

Let z = -6+8i
|z| =\(\sqrt { (-6)^{ 2 }+8^{ 2 } } \)
= \(\sqrt { 36+64 } =\sqrt { 100 } \) = 10
\(\sqrt { a+ib } =\pm \left( \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \right) \)
[Here |z| = 10, a = -6, b = 8]
\(\sqrt { -6+8i } \pm \left( \sqrt { \frac { 10-6 }{ 2 } } +i\frac { 8 }{ |8| } \sqrt { \frac { 10+6 }{ 2 } } \right) \)
= \(\pm \left( \sqrt { \frac { 4 }{ 2 } } +i\sqrt { \frac { 16 }{ 2 } } \right) \)
= \(\pm (\sqrt { 2 } +i\sqrt { 8 } )\)
= \(\\ \pm (\sqrt { 2 } +i2\sqrt { 2 } )\)
Aliter :
Square root of -6 + 8i
Let a + ib = - 6 + 8i
a = -6, b = 8
\(|z|=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\)
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
\(=\pm\left[\sqrt{\frac{10-6}{2}}+i \sqrt{\frac{10+6}{2}}\right]\)
\(=\pm[\sqrt{2}+i \quad 2 \sqrt{2}]\)

y = 4 sin (-2x)
The amplitude of sin x is 1
\(\Rightarrow\) amplitude of sin (-2x) is 1
\(\therefore\) Amplitude 4 sin(-2x) is 4 \(\times\) 1 = 4.
The period of sin \((-2x)is2x=2\pi \Rightarrow =\frac { 2\pi }{ 2 } =\pi \)

\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
A is a matrix of order (2 \(\times\) 4)
∴ \(\rho \)(A) ≤ min(2, 4) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} 1 & -2 \\ 3 & -6 \end{matrix} \right| \) = -6 + 6 = 0
Also, \(\left| \begin{matrix} -1 & 0 \\ -3 & 1 \end{matrix} \right| \) = -1 + 0 = -1 ≠ 0
∴ \(\rho \)(A) = 2

Vector form of the equation of the plane is
\(\vec { r } .(6\hat { i } +4\hat { j } -3\hat { k } )\) = 12
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(6\hat { i } +4\hat { j } -3\hat { k } )=12\)
⇒ 6x + 4y - 3z = 0
Dividing by 12, we get
\(\frac { 6x }{ 12 } +\frac { 4y }{ 12 } +\frac { 3z }{ 12 } =1\)
[\(\because \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) is the equation of the plane in intercept form]
⇒ \(\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { z }{ -4 } =1\)
∴ The x-intercepts of the plane is 2, y intercept is 3 and z-intercept is -4.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Here A = 2, B = 0, C = -1, F = -7
Here A ≠ C and A and C are of opposite signs.
Hence the given equation represents a hyperbola.

We know that the volume of the parallelepiped whose coterminus edges are \(\vec { a } ,\vec { b } ,\vec { c } \) is given by |\([\vec { a } ,\vec { b } ,\vec { c } ]\)|. Here, \(\vec { a } =\hat { 2i } -\hat { 3j } +\hat { 4k } ,\vec { b } =\hat { i } +\hat { 2j } -\hat { k } ,\vec { c } =\hat { 3i } -\hat { j } +\hat { 2k } \)
Since \([\vec { a } ,\vec { b } ,\vec { c } ]\)  = \(\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| =-7\) , the volume of the given parallelepiped is \(\left| -7 \right| =7\) cubic units.

\(\left| \frac { 2+i }{ -1+2i } \right| =\frac { \left| 2+i \right| }{ \left| -1+2i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } }{ \sqrt { \left( -1 \right) ^{ 2 }+{ 2 }^{ 2 } } } =1\) \(\left( \because \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| ,{ z }_{ 2 }\neq 0 \right) \)

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

Let X be the thrown random variable denotes the total in two the thrown a die.
Sample space S

n (S) = 36
X = {2, 4, 6, 8, 10}

\(p(x=2)=\cfrac { 1 }{ 36 } \)
\(p(x=4)=\cfrac { 4 }{ 36 } \)
\(p(x=6)=\cfrac { 10 }{ 36 } \)
\(p(x=8)=\cfrac { 12 }{ 36 } \)
\(p(x=10)=\cfrac { 9 }{ 36 } \)
(i) Probability mass function is

(ii) Cumulative distribution function .
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)(X = x_i)
P(X<2) = 0 for \(\infty\) < x < 2
\(F(2)=\frac { 1 }{ 36 } \)
\(F(4)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 36 } \)
\(F(6)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(8)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 }\)
\(F(10)=\frac { 27 }{ 36 } +\frac { 9 }{ 36 } =\frac { 36 }{ 36 } =1\)
∵ The cumulative distribution function n
\(F(x)=\left\{\begin{array}{lll} 0 & \text { for } & x<2 \\ \frac{1}{36} & \text { for } & x \leq 2 \\ \frac{5}{36} & \text { for } & x \leq 6 \\ \frac{15}{36} & \text { for } & x \leq 8 \\ 1 & \text { for } & x \leq 10 \end{array}\right.\)
(iii) p(4≤ X < 10) = p(x = 4) + p(x = 6) + p(x = 8)
= \(\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } =\frac { 13 }{ 18 } \)  
(iv) p(x ≥ 6) = p(x = 6) + p(x = 8) + p(x = 10)
= \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)
Sample space = {4 childrens}