This 2’s complement representation of(–539)10 in hexadecimal is
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Solution
Let’s go step by step:
1.539 in binary is 1000011011.
2.Taking complement to get 1’s complement we get 0111100100.
3.Adding 1 to get 2’s complement we get 0111100111.
4.Putting signed bit, i.e., 1 in MSB we get 10111100111.
5.Since, these are 11 bits and hexadecimal needs to make group of 4 bits, we have to add an extra bit.
6.Since, signed bit is 1 we pad it with 1 only in starting.
7.We get finally 110111100101.8.This is DE5 in hexadecimal.
The simultaneous equations on the Boolean variables x,y,z and w,
x +y + z =1
xy = 0
xz + w = 1
xy + \(\bar{Z}\: \bar{W}\)= 0
Have the following solution for x, y, z and w, respectively
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Solution
This is solved by heuristics and by not lengthy computations.
Since,xy= 0 and from the laste quation, we get z'w'= 0
Therefore, only option (c) satisfies both these, i.e., xy =0 and z'w' = 0
A race condition occurs when
Which of the following Query transformation (i.e.,replacing the L.H.S. expression by the R.H.S. expression) is incorrect?R1 and R2 are relations, C1 and C2 are selection conditions and A1 and A2 are attributes of R1.
A station A need to send a message consisting of 9 packet to station B using a sliding window (window size = 3) and use go back n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but not acts from B available get lost), then what is the number of packet that A will transmit for sending the message to B?
Consider the polynomial
P(x)=a0+a1x+a2x2+a3x3,where a
i ≠ 0,i
The minimum number of multiplications needed to evaluate p on an input x is _________
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Solution
The expression is a0+a1x+a2x2+a3x3
The deciding term is the last term which requires most number of multiplications.For any value of x,the computation of a3 × x × x × x is essential.Hence minimum number of multiplications is 3.
Let G (x) =\(\frac{1}{(1-x)^{2}}=\sum_{i=0}^{\infty }g(i)x^{i}\), where |x|< 1.What is g(i)?
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Solution
What are the eigenvalues of the following 2 × 2 matrix ?
\(\begin{bmatrix} 2 & -1 \\ -4 & 5 \end{bmatrix}\)-
Solution
Consider the following system of equations in three real variables x1, x2 and x3
2x1– x2+ 3x3= 1
3x1– 2x2+ 5x3= 2
–x1 + 4x2+ x3= 3
This system of equations has
Consider the set H of all 3 × 3 matrices of the type
\(\begin{bmatrix} a & f & e \\ 0 & b & d \\ 0 & 0 & c \end{bmatrix}\)where a, b, c, d, e and f are real numbers and abc ¹0.Under the matrix multiplication operation, the set H is
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Solution
(i)Set H is closed, since multiplication of upper triangular matrices will result only in upper triangular matrix.
(ii)Matrix multiplication is associative, i.e.A *(B * C) = (A *B) * C.
(iii)Identity element is
\(I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
and this belongs to H as I is an upper triangular as well as lower triangular matrix.
(iv)If A∈H, then |A| = abc.
Since abc ≠ 0,this means that |A| ≠ 0 i.e. every matrix belonging to H is non-singular and has a unique inverse.\Hence set H along with matrix multiplication is a group.