SOLUTION

Solution

P(aircraft will crash)=0.03 × 0.02 × 0.05 = 0.00003
P(air craft will not crash)=1−0.00003 = 0.99997

SOLUTION

Let the boxes be B₁, B₂, B₃, B₄
Let us assume that two specific balls have been put in box Bi
(i = 1,2,3,4). It means in box Bi
 we have to put 3 balls from the remaining 18 balls. Thus the probability that the two specific balls have been put in the particular box

Solution

Mean $$=  \mu = np  $$  = 12;  $$n =$$ number of trials ;  $$p =$$ probability of success ; $$q = (1-p) =$$ probability of failure
Variance $$ ={ \sigma  }^{ 2 } = npq  = 8 $$ 
 $$ 8 = np(1-p) = 12(1-p) $$   ; 
 $$ p = \dfrac {1 }{ 3 }  $$  
$$  \mu = np = 12$$ 
$$ n =  \dfrac { 12}{ p } = 12 \times \dfrac { 3 }{ 1 } = 36 $$ 
Parameters are $$n =36$$ and $$ p = \dfrac {1 }{ 3 }  $$

Solution

2nα = π/2

(tanα tan(2n – 1)α) (tan 2α tan(2n – 2)α) ….. (tan (n – 1)α tan (n + 1)α).tan nα  

(tanα tan(π/2 - α))(tan2αtan(π/2-2α))......tanπ/4 =1

Solution

Proceeding as usual, h=1, k=1 and the equation reduces to $$\displaystyle \frac{2+v}{\left ( 1-v \right )\left ( 1+v \right )}dv= \frac{d\alpha }{\alpha }$$
Partial Fractions $$\displaystyle \left [ \frac{3}{2\left ( 1-v \right )}+\frac{1}{2\left ( 1+v \right )} \right ]dv= \frac{d\alpha }{\alpha }.$$
Integrate $$\displaystyle -3\log \left ( 1-v \right )+\log \left ( 1+v \right )= 2\log \alpha +\log k$$ or $$\displaystyle \frac{1+v}{\left ( 1-v \right )^{3}}= k\alpha ^{2}$$ or $$\displaystyle \left ( 1+\frac{\beta }{\alpha } \right )= k\alpha ^{2}\left ( 1-\frac{\beta }{\alpha } \right )^{3}$$ or $$\displaystyle \left ( \alpha +\beta  \right )= k\left ( \alpha -\beta  \right )^{3}$$ where $$\displaystyle \alpha = x-h= x-1, \beta = y-k= y-1$$
$$\displaystyle \left ( x+y-2 \right )= k\left ( x-y \right )^{3}.$$ 

Solution

Let $$A$$ be the event that face 1 turns up
$$ B$$ be the event that face 2 turns up.
Then $$\displaystyle P\left ( A \right )= 0.10$$ and $$\displaystyle P\left ( B \right )= 0.32,$$
Since A, B are mutually exclusive, we have
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )= 0.10+0.32= 0.42.$$
We are to find $$\displaystyle P\left ( A|A\cup B \right )$$
But $$\displaystyle P\left ( A|A\cup B \right )= \frac{P\left [ A\cap \left ( A\cup B \right ) \right ]}{P\left ( A\cup B \right )}$$

$$= \dfrac{P\left ( A \right )}{P\left ( A\cup B \right )}$$

Hence $$\displaystyle P\left ( A|A\cup B \right )= \frac{0.10}{0.42}= \frac{5}{21}$$

Solution

Given equation:

x²−3?x?+2=0

When x>0, we have

x2²−3x+2=0

⇒(x−2)(x−1)=0

⇒x=2 or x=1

and when x<0,

x²+3x+2=0

or, (x+2)(x+1)=0

or, x=−2 or x=−1

Therefore, number of real roots is 4.

SOLUTION

SOLUTION
 

Let f (x) = k ( x - α ) ( x - β - i ) ( x - β + i )

= k ( x + α) ( ( x - β )² + 1 )

f' (x) = k ( ( x - β )² + 1 ) + k ( x - α ) ( 2 ( x - β ) )

⇒ f'(x) = k ( 3x² - 2 ( α + 2β ) x + ( β² + 2αβ + 1 )

? = 4 ( α + 2β )² - 4.3 ( β² + 1 + 2αβ )

? = 4 {α² + β² - 2αβ - 3 }

? = 4 { ( α - β )² - 3 }

For equal or complex roots , We have

( α - β )² - 3 ≤ 0

=|α−β|≤3√, where | α - β | = AD.

Hence maximum length of altitude AD is 3√ .