Scripted data : A data that you access using java script is termed as scripted data.
Summary Data : Summary data means a new record is created to store values in a row of a relation.

→ Here, every process requirement is 4 instances of the resource.
→ If we allocates 3 instance( one instance less than the requirement of each process) of the resource and to one process we allocate its minimum requirement then in that way with limited available instance of resource, without entering into deadlock we can fulfill requirement of maximum number of processes.
→ Now in question it is given that we have 12 instance then using above strategy we can allocate resources to 3 process without entering into deadlock.

In Readers-Writers problem, more than one Reader is allowed to read simultaneously but if a Writer is writing then no other writer or any reader can have simultaneous access to that shared object. So Writers are given exclusive access to shared objects.

BFS → O( b ^d ) worst case space complexity
DFS → O(bm) worst case space complexity
Depth- Limited Search → O(bl)
Iterative deepening Search → O(bd)
Note: Based upon BFS and DFS we can find the solution.

→ AMPS, D-AMPS, CDMA, Broadband CDMA are the standard methods used for cellular communication.
→ ​ AMPS(Advanced Mobile Phone Service):​ It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.
→ ​ D-AMPS(Digital- Advanced Mobile Phone Service) :​ It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.
→ ​ CDMA(Code Division Multiple Access) :​ In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.
→ ​ Broadband CDMA :​ Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.

→ The session layer is the network dialog controller. It establishes, maintains, and synchronizes the interaction among communicating systems.
→ ​ Dialog control:​ The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.
→ ​ Synchronization​ : The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.

if L is a regular language then we always have a string “w” which can be broken into “xy” such that |y|=2|x|
Consider a language L= {a​ 3n​ | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, .......}
then L​ ₁​ = {a^{​ n}​ | n ≥ 0} as every string in L can be broken into three equal parts .
DFA for L

DFA for L​ ₁

By same logic L​ ₂​ is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L​ ₁​ , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”

The Software Requirement Specification(SRS) is said to be Consistent if and only if no subset of individual requirements described in it conflict with each other.
Unambiguous : ​ SRS said to be unambiguous if there exist no word that have more than one meaning and this will not confuse testers to get the exact reference.
Correct : SRS is said to be correct if it is error free and accurate.

Here, number of models is nothing but number of TRUEs in final statement. In this proposition logic we got total 15 number of models.

Clean-room Strategy:
Cleanroom software engineering (CSE) is a process model that removes defects before they can precipitate serious hazards. It is a team-oriented, theory based software, which is developed using the formal methods, correctness verification and Statistical Quality Assurance (SQA).
→ Clean room management is based on the incremental model of software development, which accumulates into the final product. The approach combines mathematical-based methods of software specification, design and correctness verification with statistical, usage-based testing to certify software fitness for use.
→ The main goal of clean room engineering is to produce zero error-based software by allowing correct designs, which avoid rework.

10-bits per pixel it means we will have 2​ ¹⁰ ​ entries in color lookup table.
8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).
24-bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2^{​ 10}​ * 3) Bytes
the size of lookup table is = 3072 Bytes = (2​ ¹⁰​ + 2^{​ 11}​ ) Bytes

● A mutual exclusion (mutex) is a program object that prevents simultaneous access to a shared resource.
● This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource.
● Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts.
● When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource
● In the question, Three process are cooperating processes and satisfying mutual exclusion condition. If process Q is executing in its critical section means remaining two processes in wait stating and they won’t enter into critical section.

An instruction size is given as 32-bits.
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(2​ 6​ ) so to identify each register uniquely 6-bits are needed.
Address part : 256K(2​ ¹⁸​ ) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Operation code:
Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)
Size of Operation code= 7-bits

Waiting time = Turnaround time- Burst time
Turnaround time = Completion time - Arrival time.
Since in question Arrival time is not given so consider it as “0” for all the processes .

● The text-align property sets the horizontal alignment (like left, right, or center) of the content in or .
● The vertical-align property sets the vertical alignment (like top, bottom, or middle) of the content in or .

If we have no repetition in a string then the number of substrings can be found using the formula :
n*(n+1)/2 + 1
We have added 1 because it may include a NULL string also.
The number of substrings = 10*(11)/2 +1
The number of substrings = 56

Statement 1 is correct because firewall works on the Application layer, so it can screen the traffic going into and out of the traffic.
Statement 2 is correct.Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Propositions r and s are logically equivalent if the statement r ↔ s is a tautology. ∼q⇒ ∼p

According to above table,
equivalence means p ⇔ q
Contrapositive means p⇒ q : ∼q⇒ ∼p
Converse means p⇒ q : q⇒ p
Implication means p ⇔ q

→ L​ ₁​ is Context Free language because we can push all the a’s that come as input and when b’s come as input pop “n” number of a’s for “n” b’s from top of the stack and when again a’s come as input pop-up all the remaining a’s(i.e. “M” number of a’s) from top of the stack for “m” number of a’s coming as input.
Since we can have a pushdown automata for L​ ₁​ so we can say L​ ₁​ is a CFL.
→ L​ ₂​ is not a context free language because L is equivalent to language= {a​ ^p​ b^{​ p}​ a^{​ p}​ | p ≥ 0 }.
So for every “b” we can POP a’s came before “b” but for “a” which came after “b” we have nothing to POP on the top of stack. Since we can’t have a pushdown automata that can accept L​ ₂​ .
L​ ₂​ is not a CFL.

● An and–or tree is a graphical representation of the reduction of problems (or goals) to conjunctions and disjunctions of subproblems (or subgoals).
● A solution in an AND-OR tree is a sub tree whose leaves are included in the goal set

The homogeneous coordinates of the cartesian point (x,y) is (x,y,1) and the transformation Matrix is a 3x3 Matrix.
The image in figure ​ a ​ is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.

Sequence in which the given packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted :
Step 1: DNS query is sent to Domain name server to convert the Domain name into it’s IP address.
Step 2: TCP SYN request packet is sent by sender to establish a connection for data transmission.
Step 3: HTTP GET request is used to request data from a specified IP address.

Average access time = level 1 hit rate( level 1 access time)+ (level1 miss rate)(level 2 hit rate(level 2 access time)+ (level 1 miss rate)( level 2 miss rate) (main memory access time)
Average access time = 0.7(0.5)+ 0.3(0.8)(5)+ 0.3(0.2)(100)
Average access time = 7.55 ns

● In computers, subtraction is generally carried out by 2’s complement.
● In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value;
● In the subtraction there may possibility of negative number as a result.