# Recent posts tagged gate2019

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hi I have already posted the same thing . Please have a look ! My question is different !
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For answering there is no need to execute the query, we can directly answer this as $2$ How? Group by Student_Names It means all name that are same should be kept in one row. There are $3$ names. But in that there is a duplicate with Raj being repeated $\implies$ Raj produces ...
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Answer: (B) Bind: Binds the socket to an address Listen: Waits for connections to the socket Accept: Accepts a connection to the socket Recv: Receives data from connection From Man page of accept: It extracts the first connection request on the queue of ... descriptor referring to that socket. The newly created socket is not in the listening state. The original socket is unaffected by this call
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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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Let us define a new function $g:$ $g(y) = f(y) -f(y+1)$ Since function $f$ is continuous in $[0,2],$ $g$ would be continuous in $[0,1]$ $g(0) = -2, g(1) = 2$ Since $g$ is continuous and goes from negative to positive value in $[0,1],$ at some point $g$ would be $0$ in $(0,1).$ $g=0 \implies f(y) = f(y+1)$ for some $y \in (0,1).$ Therefore, correct answer would be $(A).$
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ans should be B. suppose E contains 2,3,4,5 and F contains 2,3,6 E U F = 2,3,4,5,6; here f(E U F) has 5 and 6 in relation... but in f(E) U f(F) does not have any relation between 5 and 6. E intesection F = 2,3; here f(E) int f(F) has same relations as f(E int F) has.
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Ans: because we have given wavy edges form MST So, for verification of option A we have to check that with MST how many cost to reach at a->b then we will get a->e->d->b = -2+5+3 = 6 so in given option a with cost(a,b)>= 6 this is posible coz , cost ... making MST so cost must be >6 not equal to 6 so option A is Need Not HOLD. Like wise if you check for other option then enequality is holding...
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see my comment on the question. That's a mistake actually. Unit productions are of type A-->B. there shouldn't be any terminal on the RHS.
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Answer - B. Multiplying $2\ 8$ bit digits will give result in maximum $16$ bits Total number of multiplications possible $= 2^8 \times 2^8$ Hence, space required $= 64K \times 16$ bits
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Consider a database with three relation instances shown below. The primary keys for the Drivers and Cars relation are did and cid respectively and the records are stored in ascending order of these primary keys as given in the tables. No indexing is available in the database. ... using primary key, then $n$ lies in the range: $36 - 40$ $44 - 48$ $60 - 64$ $100 - 104$
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How many bytes of data can be sent in $15$ seconds over a serial link with baud rate of $9600$ in asynchronous mode with odd parity and two stop bits in the frame? $10,000$ bytes $12,000$ bytes $15,000$ bytes $27,000$ bytes
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