# Recent posts tagged preparation

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@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
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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
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This is not a programing website :) But let me help you. you can redirect from your html to php page by meta tag in head <meta http-equiv="refresh" content="0; url=http://your.com/index.php" />
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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.
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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.
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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
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Passing a constant value will produce the same result in both these parameter passing mechanisms or in almost any of the parameter passing mecahinsms.
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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.
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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
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At each fork() the no. of processes becomes doubled. So, after $3$ fork calls, the total no. of processes will be $8$. Out of this $1$ is the parent process and $7$ are child processes. So, total number of child processes created is $7$.
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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) =$\left(\dfrac{1}{e^{3}}\right) + \left(\dfrac{3}{e^{3}}\right) + \left(\dfrac{9}{2e^{3}}\right)$ =$\dfrac{17}{\left(2e^{3}\right)}$
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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
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The protocol data unit (PDU) for the application layer in the Internet stack is: Segment Datagram Message Frame
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