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  • 2 answers

Madhav Kaushik 3 years ago

Hello

Gembali Harsha Vardhan 3 years, 1 month ago

Given An ice-cream parlour receives a customer at an average rate of 4 per minute. If the number of customers received by the parlour follows a Poisson distribution, To Find the approximate probability that 16 customers will be coming to the parlour in a particular 4-minute period on a given day Solution: an average rate of 4 per minute. => an average rate of 4* 4 per 4 minute. => an average rate of 16 per 4 minute. Hence mean = 16 λ = 16 = P(x) = λx e^(-^) / x! P(16) = 16¹6e-16/16! = 0.099 = 0.1
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  • 1 answers

Preeti Dabral 1 year, 11 months ago

Step 1 of 2

Two cards are drawn simultaneously,

Let, denotes the number of aces in 2 draws from a pack of 52 cards

∴ X can takes values 0, 1, 2

Step 2

{tex}\begin{aligned} & P(X=0)=\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{12 \times 47}{13 \times 51}=\frac{188}{221} \\ & \therefore P(X=0)=\frac{188}{221} \\ & P(X=1)=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 2}{52 \times 51}=\frac{32}{221} \\ & \therefore P(X=1)=\frac{32}{221} \\ & P(X=2)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \\ & \therefore P(X=2)=\frac{1}{221} \\ & \end{aligned}{/tex}

Thus the probability distribution of random variable is 
{tex}\begin{array}{|l|l|l|l|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221}= & \frac{32}{221}= & \frac{1}{221}= \\ \hline \end{array}{/tex}

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  • 1 answers

Nitasha Goyal 3 years, 4 months ago

Wrong question
  • 3 answers

Samkit Jain 3 years, 1 month ago

Not yet published as hard copy,, but available in pdf form

Siddharth Tripathi 3 years, 3 months ago

pdf is now avialable online

Sahil Gupta 3 years, 3 months ago

Not yet published by cbse
  • 0 answers
  • 2 answers

Sahil Gupta 3 years, 3 months ago

Ruby Rawat on unacademy

Arnob Phukon 3 years, 5 months ago

Search ..Applied mathematics Gaur Classes
  • 2 answers

Tanushri Bhayani 1 year, 8 months ago

Yes partnership and index number

Sourav Khandelwal 3 years, 5 months ago

No
  • 0 answers
  • 0 answers
  • 2 answers

Sher Singh 3 years, 3 months ago

10

Anamika Sharma 3 years, 7 months ago

10
  • 0 answers
  • 5 answers

Arnob Phukon 3 years, 5 months ago

ML Agarwal

Kajal Kumari 3 years, 6 months ago

Rs Agarwal is also good according to me

Prachi Janwani 3 years, 7 months ago

Rd sharma is also good according to me

Harman Singh 3 years, 7 months ago

dont ever use ML agarwal, its the worst book and very complicated, i used it in class 11th

Piyush Singh 3 years, 7 months ago

Ml Agarwal
  • 1 answers

Rakamanickam Meerabai 3 years, 7 months ago

f'(x)=-7 <0 Therefore f(x) is strictly decreasing.
  • 0 answers
  • 1 answers

Preeti Dabral 1 year, 11 months ago

Linear programming deals with the optimization (maximization or minimization) of a linear function of a number of variables subject to a number of conditions on the variables, in the form of linear inequations or equations in variables involved.

  • 5 answers

Sumit Singh Bora 3 years, 9 months ago

Vishal Kumar 3 years, 9 months ago

2x

Hrithik Chauhan 3 years, 9 months ago

x^2

Tanay Bhushan 3 years, 11 months ago

Tanmay Jain 3 years, 11 months ago

X2
  • 1 answers

Golden K 3 years, 11 months ago

[(X^4) /4 - (X^2) /2] from 1 to 2 =9/4
  • 1 answers

Arjűņ _Pāwåř 4 years, 2 months ago

This is from laplace transform- L{sinat}=s2s2+a2[−e−st(1ssinat+as2cosat)]t=∞t=0L{sin⁡at}=s2s2+a2[−e−st(1ssin⁡at+as2cos⁡at)]t=0t=∞ L{sinat}=∫→+∞0e−stsinatdtL{sin⁡at}=∫0→+∞e−stsin⁡atdt I=∫→+∞0e−stsinat=12i∫→+∞0e−st(eiat−e−iat)dt=12i∫→+∞0(e−t(−ia+s)−e−t(ia+s)dtI=∫0→+∞e−stsin⁡at=12i∫0→+∞e−st(eiat−e−iat)dt=12i∫0→+∞(e−t(−ia+s)−e−t(ia+s)dt 2iI=∣∣∣e−t(−ia+s)(ia−s)∣∣∣∞0+∣∣∣e−t(ia+s)(ia+s)∣∣∣∞02iI=|e−t(−ia+s)(ia−s)|0∞+|e−t(ia+s)(ia+s)|0∞ 2iI=1(s−ia)−1(s+ia)2iI=1(s−ia)−1(s+ia) I=as2+a2 May be it will help you.. I did as i can... Thanks!
  • 1 answers

Dj King 4 years, 2 months ago

Afternoon
  • 3 answers

Rakamanickam Meerabai 3 years, 7 months ago

Let cos-1 8/17=y cos y = 8/17. sec y=17/8 tan y =√(sec^2 y--1) =√(289/64--1) = 15 /8 Therefore. y=tan-1(15/8) tan(tan-1(15/8) ) =15/8

Harsh M 3 years, 10 months ago

15/8

Tanmay Sharma 4 years, 3 months ago

15/8
  • 5 answers

Lokendra Tomar 3 years, 11 months ago

Aap kha se ho

Ravindra Bijarniya 3 years, 11 months ago

Oo bhai kaha se laya hai

Satyam Mishra 3 years, 11 months ago

Wffnt qeeeey136677#^ vevvvvvw%,'%++!%_€£*^:@:#,$?*$^#)×₩×₹£÷€÷_$&;#;÷:"×€£》~■~{○{○[》~《○■○■▪︎<>《¤☆¡▪︎¿》¤♧¤◇•]~{\>|>|[~♤○》¤》¤

Yash Sharma 3 years, 11 months ago

??oo bhai kya hai ye

Dhruv Sarkar 4 years, 5 months ago

The equation can be written as (x-1)^2 + (y+2)^2 = 5. —(1). => center of the circle is at O(1,-2). Now solve for eq(1) and y=0 so that we get the point(s) of intersection of the circle with the x-axis.The points will be A(3,0) and B(-1,0). The equation of the line OA will then be (by using two point form), x-y-3=0, which is the equation of one of the diameters,
  • 4 answers

Sher Singh 3 years, 3 months ago

Yes

Nikki Bhati 3 years, 8 months ago

Yes

Harsh M 3 years, 10 months ago

YES

Ramakant Singh 4 years, 5 months ago

yes sir
  • 1 answers

Rohit Jain 4 years, 5 months ago

How to read applied Maths for session 2020-2021

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