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1
@arjun sir is fully correct now I am going to discuss about why frame transmission time must be >= 2 * prop. Time In csma/cd the sender doesn't holds a copy after it has sent a frame so the sender sends a frame now if a collision is detected it resends after ... If the sender is still transmitting then only it can resend it else not. That's reason why frame transmission time must be > 2 prop. Time
posted Sep 1, 2020 in Others Navneet Singh Tomar 6,398 views
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Answer - B. Degree of generator polynomial is $3$ hence $3-bits$ are appended before performing division After performing division using $2$'$s$ complement arithmetic remainder is $011$ The remainder is appended to original data bits and we get $M' = 11001001\bf{011}$ from $M = 11001001.$ Courtesy, Anurag Pandey
posted Mar 16, 2019 in Preparation Experience Parth Shah 2,588 views
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At least 2 out of n people born on April 1 means either all n are not born on April 1 or exactly 1 born on April 1. So, P(X) = 1 - P(Y) - P(Z) where P(Y) is the probability that none are born on April 1 and P(Z) is the probability that exactly 1 is born on April 1 P(Y) = 364n/365n P( ... ) = 0.016 P(50) = 0.24 P(120) = 0.517 P(119) = 0.514 P(115) = 0.502 P(114) = 0.497 So, 115 would be the answer.
posted May 30, 2018 in From GO Admins Arjun 1,838 views
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Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
posted May 4, 2018 in Preparation Advice Shikha Mallick 1,126 views
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Consider a simple connected graph $G$ with $n$ vertices and $n$ edges $(n > 2)$. Then, which of the following statements are true? $G$ has no cycles The graph obtained by removing any edge from $G$ is not connected $G$ has at least one cycle The graph obtained by removing any two edges from $G$ is not connected None of the above
posted Oct 8, 2017 in Preparation Experience prateekkmr60 4,903 views
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Option B is the most correct answer. (To know why it is not THE correct answer read the tail section of this answer) The reasons are as follows. A problem p in NP is NP-complete if every other problem in NP can be transformed into p in ... The answer is partially correct because when we say reduction, we usually mean polynomial time reduction, but neverthless it doesnt hurt to be precise.
posted Jun 20, 2016 in Preparation Advice Tauhin Gangwar 931 views
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Yes. Both L and L' are not RE. We can have the same reduction as done for L(M) is infinite. Lets assume L is RE. So, we have a TM N which will say "yes" if given an encoding of a TM whose language is $\Sigma ^*$. Now using N we try to solve ... not RE as per Rice's theorem second part. So, both L and L' are not RE making (D) the correct answer. http://gatecse.in/wiki/Rice%27s_Theorem_with_Examples
posted May 15, 2016 in Others Desert_Warrior 1,140 views
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Answer is (C) Poisson Probability Density Function (with mean $\lambda$) =$\dfrac{ \lambda^{k}} { \left(e^{\lambda}k!\right)}$, We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda = 3 $(thus finding the cumulative mass function) =$\left(\dfrac{1}{e^{3}}\right) + \left(\dfrac{3}{e^{3}}\right) + \left(\dfrac{9}{2e^{3}}\right)$ =$\dfrac{17}{\left(2e^{3}\right)}$
posted Apr 2, 2016 in Preparation Experience Akash Kanase 11,869 views
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Register renaming is done in pipelined processors: as an alternative to register allocation at compile time for efficient access to function parameters and local variables to handle certain kinds of hazards as part of address translation
posted Mar 19, 2016 in Study Materials Arjun 1,845 views
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