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Yeah why not you have found a solution it cab be one of the option but always choose the best one
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Answer: D 3NF will be adequate for normal relational database design since 3NF tables are free of insertion, update, and deletion anomalies.
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more details plzzz
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Link Utilization $=\dfrac{\text{Amount of data sent}}{\text{Max. amount of data that could be sent}}$ Let $x$ ... $L=40\times 64\ bits=40\times \dfrac{64}{8}\ bytes=40\times 8\ bytes=320\text{ bytes (answer)}$
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The total number of nodes accessed including root will be 5. The order is, (9)-->(5)-->(5,7)-->(9,11)-->(13,15).
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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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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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Let $G$ be a weighted undirected graph and e be an edge with maximum weight in $G$. Suppose there is a minimum weight spanning tree in $G$ containing the edge $e$. Which of the following statements is always TRUE? There exists a cutset in $G$ having all edges of maximum ... in $G$ having all edges of maximum weight. Edge $e$ cannot be contained in a cycle. All edges in $G$ have the same weight.
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Time without pipeline $=6 \text{ stages}=6 \text{ cycles}$ Time with pipeline $=1+\text{stall freqency}\times \text{stall cycle}$ $=1+.25\times 2$ $=1.5$ Speed up $=\dfrac{6}{1.5}=4$
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answer - D AB + C = ((AB + C)')' = ((AB)'C')' = ((A' + B')C')' = (A' + B')' + C two NOR gates for complementing A and B one for computing (A' + B')' one to compute ((A' + B')' + C)' one to negate the last result
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$P\left(x\right) = \left(\neg \left(x=1\right)\wedge \forall y \left(\exists z\left(x=y*z\right)\right) \Rightarrow \left(y=x\right) \vee \left(y=1\right) \right)$
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The formula used to compute an approximation for the second derivative of a function $f$ at a point $X_0$ is $\dfrac{f(x_0 +h) + f(x_0 – h)}{2}$ $\dfrac{f(x_0 +h) - f(x_0 – h)}{2h}$ $\dfrac{f(x_0 +h) + 2f(x_0) + f(x_0 – h)}{h^2}$ $\dfrac{f(x_0 +h) - 2f(x_0) + f(x_0 – h)}{h^2}$
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Consider the following program segment for concurrent processing using semaphore operators $P$ and $V$ for synchronization. Draw the precedence graph for the statements $S_1$ to $S_9$. var a,b,c,d,e,f,g,h,i,j,k : semaphore; begin cobegin begin S1; V(a); V(b) end; begin P(a); S2; V(c); V(d) end; ... end; begin P(g); S6; V(i) end; begin P(h); P(i); S8; V(j) end; begin P(j); P(k); S9 end; coend end;
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2011 is the answer. In C, there is a rule that whatever character code be used by the compiler, codes of all alphabets and digits must be in order. So, if character code of '$A$' is $x$, then for '$B$' it must be $x+1$. Now %s means printf takes ... gives the same result as first argument to printf is a character pointer and only if we want to pass more arguments we need to use a format string.
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In order traversal of $b$ binary search tree returns the element in sorted order - ascending (inorder is left parent then right. in a bst left is less than parent and right is greater than parent). In this option $A$ is the only sorted list. hence it is the only possibility.
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One of the header fields in an IP datagram is the Time-to-Live (TTL) field. Which of the following statements best explains the need for this field? It can be used to prioritize packets. It can be used to reduce delays. It can be used to optimize throughput. It can be used to prevent packet looping.
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An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: $245.248.128.0/20$. The ISP wants to give half of this chunk of addresses to Organization $A$, and a quarter to Organization $B$, while retaining the remaining with itself. Which of the ... $245.248.136.0/24 \text{ and } 245.248.132.0/21$
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Type checking is normally done during lexical analysis syntax analysis syntax directed translation code optimization
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A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is $\dfrac{1}{6}$ $\dfrac{3}{8}$ $\dfrac{1}{8}$ $\dfrac{1}{2}$
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A $5$ stage pipelined CPU has the following sequence of stages: IF - instruction fetch from instruction memory RD - Instruction decode and register read EX - Execute: ALU operation for data and address computation MA - Data memory access - for write access, the register read ... clock cycles taken to complete the above sequence of instructions starting from the fetch of $I_1$? $8$ $10$ $12$ $15$
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Consider the binary relation: $S= \left\{\left(x, y\right) \mid y=x+1 \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$ The reflexive transitive closure is $S$ is $\left\{\left(x, y\right) \mid y >x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$ ... $\left\{\left(x, y\right) \mid y \leq x \text{ and } x, y \in \left\{0, 1, 2\right\} \right\}$
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A computer uses $46-bit$ virtual address, $32-bit$ physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table $(T1)$, which occupies exactly one page. Each entry of $T1$ stores the base address of a page of ... cache block size is $64$ bytes. What is the size of a page in $KB$ in this computer? $2$ $4$ $8$ $16$