Login
Register
Dark Mode
Brightness
Profile
Edit Profile
Messages
My favorites
My Updates
Logout
Filter
Profile
Wall
Recent activity
All questions
All answers
Exams Taken
All Blogs
Recent activity by Shashwat21225
4
answers
1
GATE CSE 2014 Set 2 | Question: 1
The security system at an IT office is composed of $10$ computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is ... are working. Let the probability that the system is deemed functional be denoted by $p.$ Then $100p =$ _____________.
The security system at an IT office is composed of $10$ computers of which exactly four are working. To check whether the system is functional, the officials inspect four...
11.9k
views
comment reshown
Jan 18, 2020
Probability
gatecse-2014-set2
probability
numerical-answers
normal
+
–
8
answers
2
GATE CSE 2014 Set 2 | Question: 4
If the matrix $A$ is such that $A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$ then the determinant of $A$ is equal to ______.
If the matrix $A$ is such that $$A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$$ then the determinant of $A$ is equal to ______.
12.9k
views
commented
Jan 16, 2020
Linear Algebra
gatecse-2014-set2
linear-algebra
numerical-answers
easy
determinant
+
–
3
answers
3
TIFR CSE 2011 | Part A | Question: 12
The action for this problem takes place in an island of Knights and Knaves, where Knights always make true statements and Knaves always make false statements and everybody is either a Knight or a Knave. Two friends A and B lives in a house. The census ... a Knave. A is a Knave and B is a Knight. Both are Knaves. Both are Knights. No conclusion can be drawn.
The action for this problem takes place in an island of Knights and Knaves, where Knights always make true statements and Knaves always make false statements and everybod...
2.1k
views
commented
Jan 14, 2020
Mathematical Logic
tifr2011
mathematical-logic
propositional-logic
+
–
4
answers
4
GATE CSE 1992 | Question: 5-a
The access times of the main memory and the Cache memory, in a computer system, are $500$ n sec and $50$ nsec, respectively. It is estimated that $80\%$ of the main memory request are for read the rest for write. The hit ratio for ... policy (where both main and cache memories are updated simultaneously) is used. Determine the average time of the main memory (in ns).
The access times of the main memory and the Cache memory, in a computer system, are $500$ n sec and $50$ nsec, respectively. It is estimated that $80\%$ of the main memor...
24.0k
views
commented
Jan 4, 2020
CO and Architecture
gate1992
co-and-architecture
cache-memory
normal
numerical-answers
+
–
2
answers
5
GATE CSE 1995 | Question: 15-a
Implement a circuit having the following output expression using an inverter and a nand gate $Z=\overline{A} + \overline{B} +C$
Implement a circuit having the following output expression using an inverter and a nand gate $$Z=\overline{A} + \overline{B} +C$$
2.6k
views
commented
Dec 30, 2019
Digital Logic
gate1995
digital-logic
k-map
normal
descriptive
+
–
1
answer
6
GATE CSE 1996 | Question: 12
Given below are the transition diagrams for two finite state machines $M_1$ and $M_2$ recognizing languages $L_1$ and $L_2$ respectively. Display the transition diagram for a machine that recognizes $L_1.L_2$, obtained from transition diagrams for $M_1$ ... $\varepsilon$ transitions and no new states. (Final states are enclosed in double circles).
Given below are the transition diagrams for two finite state machines $M_1$ and $M_2$ recognizing languages $L_1$ and $L_2$ respectively.Display the transition diagram fo...
8.5k
views
commented
Dec 24, 2019
Theory of Computation
gate1996
theory-of-computation
finite-automata
normal
descriptive
+
–
1
answer
7
GATE CSE 2014 Set 1 | Question: 17
Which one of the following is FALSE? A basic block is a sequence of instructions where control enters the sequence at the beginning and exits at the end. Available expression analysis can be used for common subexpression elimination. Live variable ... be used for dead code elimination. $x=4*5 \Rightarrow x=20$ is an example of common subexpression elimination.
Which one of the following is FALSE?A basic block is a sequence of instructions where control enters the sequence at the beginning and exits at the end. Available express...
11.8k
views
commented
Dec 17, 2019
Compiler Design
gatecse-2014-set1
compiler-design
code-optimization
normal
+
–
2
answers
8
GATE CSE 1996 | Question: 7
A demand paged virtual memory system uses $16$ bit virtual address, page size of $256$ bytes, and has $1$ Kbyte of main memory. $\text{LRU}$ page replacement is implemented using the list, whose current status (page number is decimal) is For each ... indicate the new status of the list page faults, if any, and page replacements, if any.
A demand paged virtual memory system uses $16$ bit virtual address, page size of $256$ bytes, and has $1$ Kbyte of main memory. $\text{LRU}$ page replacement is implement...
7.9k
views
commented
Dec 15, 2019
Operating System
gate1996
operating-system
virtual-memory
normal
descriptive
+
–
6
answers
9
GATE CSE 1998 | Question: 7-b
In a computer system where the best-fit' algorithm is used for allocating jobs' to memory partitions', the following situation was encountered:$\begin{array}{|l|l|} \hline \textbf{Partitions size in $KB$} & \textbf{$ ... $} \\\hline \end{array}$When will the $20K$ job complete?
In a computer system where the ‘best-fit’ algorithm is used for allocating ‘jobs’ to ‘memory partitions’, the following situation was encountered:$$\begin{arr...
12.9k
views
commented
Dec 11, 2019
Operating System
gate1998
operating-system
process-scheduling
normal
+
–
1
answer
10
SELF_DOUBT
i have a line in my notes-------- IN Order to switch from User mode to Kernel mode during execution . we just have to change one Bit at hardware level.. I am not remembering concept plzz help
i have a line in my notes IN Order to switch from User mode to Kernel mode during execution . we just have to change one Bit at hardware level.. I am not remember...
366
views
answered
Dec 7, 2019
Email or Username
Show
Hide
Password
I forgot my password
Remember
Log in
Register