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Maths-

General

Easy

Question

Three numbers are chosen at random from the set {1,2,3,.....20} The probability that the minimum of these chosen numbers is 7 is

$\frac{13}{190}$
$\frac{26}{57}$
$\frac{11}{190}$
none of these

The correct answer is: $fraction numerator 13 over denominator 190 end fraction$

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An integer is selected at random from between 1 and 200, then probability that integer is divisible by 2, 3, 6, or 8 is

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In the above problem the density of liquid is

In the above problem the density of liquid is

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A piece of wood of mass 150g has a volume of 200 cm³ Then the density of wood in C.G.S system is

A piece of wood of mass 150g has a volume of 200 cm³ Then the density of wood in C.G.S system is

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Which one of these is a derived physical quantity?

Which one of these is a derived physical quantity?

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Which of the following numbers is a perfect number?

Which of the following numbers is a perfect number?

Which of the following is a perfect number?

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parallel

$c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$c o s invisible function application e c to the power of 2 end exponent 60 to the power of ring operator end exponent s e c to the power of 2 end exponent invisible function application 30 to the power of ring operator end exponent c o s invisible function application 0 to the power of ring operator end exponent s i n invisible function application 45 to the power of ring operator end exponent c o t to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

$4 open parentheses s i n to the power of 4 end exponent invisible function application 30 to the power of ring operator end exponent plus c o s to the power of 4 end exponent invisible function application 60 to the power of ring operator end exponent close parentheses plus fraction numerator 2 over denominator 3 end fraction open parentheses s i n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent minus c o s to the power of 2 end exponent invisible function application 45 to the power of ring operator end exponent close parentheses plus fraction numerator 1 over denominator 2 end fraction t a n to the power of 2 end exponent invisible function application 60 to the power of ring operator end exponent equals$

cot $theta$ + tan $theta$ =

Hence option 2 is correct

cot $theta$ + tan $theta$ =

Hence option 2 is correct

parallel

$open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals$

Hence option 2 is correct

$open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals$

Hence option 2 is correct

If $s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction$ , then cos A cosecA + tan A sec A =

If $s i n invisible function application A equals fraction numerator 1 over denominator 3 end fraction$ , then cos A cosecA + tan A sec A =

If sin B = 1/2 then 3cos B – 4 cos³ B =

Hence option 4 is correct

If sin B = 1/2 then 3cos B – 4 cos³ B =

Hence option 4 is correct

parallel

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