________If your program compiles successfully, it will produce correct results. (T/F)

Answer:

if(y==10)

{

     x=0;   // assigning 0 to x if y equals 10.

}

else

{

   x=1;   // assigning 1 to x otherwise.

}

Explanation:

In the if statement i have used equal operator == which returns true if value to it's right is equal to value to it's left otherwise false.By using this operator checking value of y and if it is 10 assigning 0 to x and if it is false assigning 1 to x.

Answer:

False.

Explanation:

A computer program can not only constructed in a high level language but also in low level language such as assembly language used for microprocessor for ex-8085 microprocessor,8086 microprocessor ,ARM assembly language and there are many more.

High level languages are programmer friendly they are easily understood by the programmers because they uses english words while low level languages are difficult because they uses binary language 0's or 1's.

Hence the answer to this question is False.

Answer:

float area;

Explanation:

The float datatype is used for storing the decimal number .The syntax of declaring float variable is given below.

float variablename ;

float area;

area=89.900002; // storing the value

following the program in c language

#include<stdio.h> // header file

int main() // main function

{

  float area=89.900002; // declared and initialize a float variable

  printf("%f",area); // display the area value

return 0;

}

Answer:

All the individual programming language have their individual important which is specific according to the their own style of leaning the programming language and simplicity to learn. The syntax should be in proper format and follow legacy.

Some of the benefit of the single programming language for all the domain are as follows:

Answer: Normalization is known as the process under which one organizes a database in order to improve data integrity and reduce redundancy.  It also tends to streamlines database design in order to achieve optimal structure made up of the basic elements.

It is also referred to as data normalization, it is considered as a vital part of database design, since it helps with accuracy, speed and efficiency of database.  By doing so, one can arrange data into columns and tables.

Answer: IPv4(Internet protocol version 4)) =32 bits

               IPv6(Internet protocol version 6) =128 bits

Explanation: IP(Internet protocol)address are the unique addresses that is provided to the hardware components that are attached with the internet service. The identification of the hardware devices are made on the basis of IP address .It is a numerical notation .

The most commonly used version of the IP addresses are IPv4 (Internet protocol version 4)and IPv6(Internet protocol version 6) that persist of 32 bits and 128 bits address respectively. IPv4 and IPv6 has different bits due to the growth of the internet facility and improvement in the version.Most commonly used internet service at current time is IPv6.

Answer: (D) MAC address

Explanation:

  MAC address is the media access control address that basically used as identification number in the hardware. It is uniquely identified in the network for the every devices.

Whenever the packet moves from source to the destination the MAC address changes and updated with new destination and source MAC address. But it does not change its IP address during this process.

Therefore, MAC address basically change with the connection when there is data transverse in the network.

Answer:

We have used int datatype which is wrong in the given declaration.

The correct declaration is public static final double myNumber = 17.36;

Explanation:

In this declaration the value is stored in "myNumber" variable is 17.36 which is decimal point number to store the decimal point number we have to use the double datatype, not int datatype because int datatype can store only integer value not the decimal value. So we have to use the double datatype instead of int type.

The correct syntax for the following declaration is :

public static final double myNumber = 17.36;

Answer:RISC(reduced instruction set computer) is the computer used for low level operation by simple command splited into numerous instructions in single clock only and CISC(Complex instruction set computer) is the computer that performs the operation in single instruction.

RISC architecture has hardwired control unit,data cache unit,data path ,instruction cache and main memory as components .CISC architecture persist of  control unit,micro program control memory, cache, instruction and data path and main memory.

The differences between RISC and CISC are as follows:-

RISC (Reduced Instruction Set Computing) architectures use simpler instructions for enhanced speed and efficiency, while CISC (Complex Instruction Set Computing) architectures use complex instructions to reduce program instruction counts. The comparison highlights differences in execution speed, pipeline efficiency, memory usage, and design philosophy. Each architecture offers unique benefits depending on application requirements.

RISC (Reduced Instruction Set Computing)

RISC architectures are designed to execute a small and highly optimized set of instructions. This simplicity enables faster instruction execution because each instruction is executed in a single clock cycle. Key features of RISC include:

An example of a RISC architecture is the MIPS (Microprocessor without Interlocked Pipelined Stages) architecture, which uses 32-bit instructions and multiple instruction formats, adheres to the principles of simplicity and efficiency.

CISC (Complex Instruction Set Computing)

CISC architectures, on the other hand, aim to reduce the number of instructions per program by using complex and multifunctional instructions. These instructions can execute multiple operations or address modes within a single instruction. Key features of CISC include:

An example of a CISC architecture is the Intel x86 family, widely used in desktop and server processors.

Comparison of RISC and CISC

To compare both architectures:

Answer:(d) MyISAM

              e) InnoDB

Explanation: MyISAM is type of default engine storage in table form in MySQL 5.5.It is discontinued and no longer available service.It had the feature of creating a backup, able to create replicas,data dictionary automatic updates etc in table.

InnoDB took over on the place of MyISAM in MySQL 5.5 as the default engine storage when MyISAM was discarded. It has the capability to support foreign keys, tablespaces, operation of spatial nature etc.

Other options are incorrect because black-hole is mechanism of data packet destruction without the acknowledgement of sender, supernova is type of computer unit and memory is the operating unit part where the data storage takes place.Thus the correct options are (d) and (e).

Answer:c)Predictive analytic

Explanation: Predictive analytics is the tool that is used for the prediction about the future situations.It is the advanced branch of the analytics where the present information considered for prediction The techniques followed by the predictive analytics are data mining, modelling etc.

Other option are incorrect because DSS(decision support system) is the system that helps in making decision in business field, reporting shows the low efficiency or performance and queries are the question or objection that arises regarding a concept.Thus the correct option is option(c).

Answer:

//Here is PSEUDO CODE.

1. Declare variable "hour_day".

   1.1 read the hour worked in a day and assign it to variable "hour_day".

2. Calculate the hours worked in a five day week.

   2.1 work_week=5*hour_day.

3. Calculate the hours worked in 252 days year.

   3.1 work_year=252*hour_day

4.  print the value of work_week and work_year.

5.  End the program.

//Here is code in c++.

#include <bits/stdc++.h>

using namespace std;

int main() {

double hour;

cout<<"enter the hours worked in a day:";

cin>>hour;

double work_week=5*hour;

double work_year=252*hour;

cout<<"work in a 5-day week: "<<work_week<<"hours"<<endl;

cout<<"work in a 252-day year: "<<work_year<<"hours"<<endl;

return 0;

}

Output:

enter the hours worked in a day:6.5

work in a 5-day week: 32.5 hours

work in a 252-day year: 1638 hours

Answer: Local area network(LAN) is the network works in the small area to create networking pattern such as an individual building. Wide area network(WAN) is the networking technology that covers a large geographical area to connect them such as any particular company.

MAN (metropolitan area network) is the area that falls in between the category of LAN and WAN.It is the networking pattern that can work in the geographical areas like towns,cities etc.WAN technology consist of both LAN and MAN networks to make a huge unit of network.Thus, MAN is the middle networking technology as it covers more area than LAN and Less area than WAN.

Final answer:

The variable t will store true for statements A and D because the comparisons are correct, and false for statements B and C because the comparisons are incorrect.

Explanation:

The question is asking about the values that will be stored in the variable t after each given boolean (true or false) statement is executed in a programming context. Here are the answers:

A) t = (12 > 1); The statement is true because 12 is greater than 1, so t will store true.

B) t = (2 < 0); The statement is false because 2 is not less than 0, so t will store false.

C) t = (5 == (3 * 2)); The statement is false because 5 is not equal to 6 (3 times 2), so t will store false.

D) t = (5 == 5); The statement is true because 5 is equal to 5, so t will store true.

Answer:

All elements in an array have the same subscript.

Explanation:

The statement written above is false because every element in an array has different subscript or index.The index starts with 0 and ends at size -1 .You can access the array element if you write the subscript in the square brackets after the name of the array.

a[5];

You can access 6th element in the array by this statement.

Answer: True

Explanation: CISO(Chief information security officer) is the officer that is responsible for the information security(Infosec) function to be implemented in the organization.He/she manages the procedures, protect from the threats in business, secures the enterprise communication.

The post of CISO is considered as a senior level officer .The responsibilities of the CISO also includes getting along with other section of the organization and assuring about the smooth and secured processing of the business work as the aspect of information security.Thus the statement given is true

Answer: 2) 802.2

Explanation: Ethernet standard are the standard for the networking technology for the transmission of the data.The ethernet is requires for the node connection and then forming the framework for the transmission of information.There are several standards for Ethernet.

802.3 is Ethernet standard that works on 10 Mbps rate, PPP(point to point protocol) is the standard that connects the internet service provider using modem and LLC(logical link control) is a standard for 802 .There is no 802.2 standard present in the ethernet standard.Thus,the correct option is option(2).

Answer:

True.

Explanation:

When creating a database it should be carefully planned so that we can easily do data manipulation and report generation if not designed carefully these processes will become difficult.

Some mistakes in the creation of database.

Answer:

The resultants bits for two strings are 0111 1110

Explanation:

Given details:

two pair of strings are

10010100

11101010

comparing two pair of strings, then perform XOR operation

Rules to follow for XOR

If both the bits are same then resultant is zero 0 and if bits are different resultant is one

XOR operation:

10010100

11 1 01010

-------------

01111110

The resultants bits for two strings are 0111 1110

Answer:

Destructor:-To free the resources when lifetime of an object ends.

Accessor:-To access private properties of the class.

Explanation:

The purpose of a destructor is to free the resources of the object that it has acquired during it's lifetime. A destructor is called once in the lifetime of the object that is also when  the ifetime of the object ends.

The purpose of the accessors is to access private properties of the class.Accessor cannot change the value of the private members.They are also called getters.

Answer:

Different business or firm tend to enforce different Information Systems based completely on their main business operations, in order to best leverage data as an organizations asset. Some of these Information System are as follow:

a) Transaction Processing System (TPS):

A small organization tends to process transactions that might result from day-to-day activities, such as purchase orders, creation of paychecks and thus require using TPS.

b) Management Information System(MIS):

Managers and owners of small organizations tend to incline towards industry-specific MIS, in order to get historical and current operational data, such as inventories data and sales.

c) Decision Support System (DSS):

A DSS to allow managers and owners of small organizations to use predefined report in order to support problem-resolution decisions and operations planning.

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

Answer:

/*

Problem; to finf s2 strings are anagrams.

(s2 strings are anagrams if they contain exactly the same letters, ignoring capitalization, punctuation,)

and spaces.

Solution: we must take account wich letters are present in each string and how many times. To achieve that we

may create a struct with a char and a char count, some constructors, both geters and an equal operator.

*/

#include <iostream>

#include <vector>

using namespace std;

struct charCount {

private:                                                                    // Internal data structure:

   char letter;                                                            // letter (should be unique)

   unsigned int letterCount;                                               // and is present at least 1 time    

public:

   charCount() {}                                                          // Default Constructor

   charCount(char l) {letter = l; letterCount = 1;}                        // Constructor

   charCount(const charCount & other) {letter = other.letter;

                                       letterCount = other.letterCount;}   // Copy Constructor

   void increments() {letterCount++;}                                      // Counts other occurrence

   char ltr() {return(letter);}                                            // Geter

   unsigned int ltrCount() {return(letterCount);}                          // Geter

   bool operator ==(charCount other) {return(letter == other.letter        // Equal means same char and

                                     && letterCount == other.letterCount);}// same count.

};

/*

We have to analyze each input string returning an ordered vector of charCount elements:

*/

vector<charCount> analysis(string s) {                                // Raw string data

vector<charCount> result;                                             // Result vector

for(char c:s) {                                                       // For each char of string: we take a-z

   if(isalpha(c)) {                                                  // only into account and we normalize it

       char x = tolower(c);                                          // to lowercase, store as x and we are

       bool found = false;                                           // ready to search

       unsigned int pos = 0;                                         // from first position to last

       for(auto & y:result) {                                        // for each vector's element

           if(y.ltr() == x) {                                        // if we find x, we must

               y.increments();                                       // count this occurrence,

               found = true;                                         // turn on the found flag

               break;                                                // and exit from the loop.

               } else if(y.ltr() < x) {                              // Otherwise may be we are not going to

               result.emplace(result.begin()+(pos),(charCount(x)));  // find it in the ordered vector: we insert

               found = true;                                         // it at current vector's position. It is

               break;                                                // present now: we should exit from loop.

               } else pos++;                                         // Otherwise we increment pos and go on.

           }

       if(!found) {                                                  // x doesn't belongs to vector's set: we

           auto it = result.end();                                   // add it at last position, keeping order

           result.insert(it,(charCount(x)));                         // and we may go on for another cycle if any.

           }

       }

   }

return result;

}

template <typename T>                                           // Generic code follows to implement

bool operator ==(vector<T> v1,vector<T> v2) {                   // 2 vectors comparison: they are

bool result = (v1.size() == v2.size());                         // equals if their sizes match,

if(result)                                                      // and each element from first vector

   for(unsigned int i = 0; i < v1.size();i++)                  // matches with each element from the

       result = result && v1[i] == v2[i];                      // second vector at the same position.

return(result);

}

template bool operator ==(vector<charCount>,vector<charCount>); // Orders compiler to generate a charCount version.

bool areAnagrams(string s1,string s2) {                         // Anagrams implies 2 identical analysis vectors.

return(analysis(s1) == analysis(s2));}

int main(int argc, char *argv[]) {                                      // Test code folloes:

string s1,s2;                                                           // 2 strings to store input data.

cout << "Text 1:"; getline(cin,s1);                                     // Reads string 1 till <Enter> key.

cout << "Text 2:"; getline(cin,s2);                                     // Reads string 2 till <Enter> key.

cout << (areAnagrams(s1,s2)?"Are":"Are not") << " anagrams..." << endl; // Are they? We use the ternary operator.

return 0;

}

Explanation: included inside the code as comments.

Answer:

#include <iostream>

using namespace std;

void areAnagrams(char str1[], char str2[])

{

   int i, flag = 1,  count1[26] = {0}, count2[26] = {0};

   for(i = 0; str1[i] != '\0'; i++) {

       if (isalpha(str1[i])) {

           str1[i] = tolower(str1[i]);

           count1[str1[i] - 'a']++;

       }

   }

   for(i = 0; str2[i] != '\0'; i++) {

       if (isalpha(str2[i])) {

           str2[i] = tolower(str2[i]);

           count2[str2[i] - 'a']++;

       }

   }    

   for(i = 0; i < 26; i++) {

       if (count1[i] != count2[i])

           flag = 0;

   }

   if (flag == 0)

       cout << "The two strings are NOT anagrams";

   else

       cout << "The two strings are anagrams";

}

int main ()

{

   char str1[100], str2[100];

   cout << "Enter the first string: ";

   gets(str1);

   cout << "Enter the second string: ";

   gets(str2);

   areAnagrams(str1, str2);

   return 0;

}

Explanation:

Inside the function:

- In the first loop, each character of the first string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.

- In the second loop, each character of the second string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.

- In the third for loop, compare the frequencies

- If they are equal, that means the strings are anagrams

Answer: The advantage of the optical fibre over the other medium of the communication are as follows:-

Optical fibres have significant technical advantages including low loss, allowing for longer distances without signal amplification, high bandwidth supporting more data transmission, and reduced crosstalk for clearer communication over traditional copper cables.

Optical fibres offer significant advantages over traditional communication media like copper cables, especially for long-distance communications. One of the primary technical advantages of optical fibres is their low loss, allowing light to travel many kilometers before needing amplification, which is much superior to the performance of copper conductors. Another critical advantage is high bandwidth, where lasers emit light that enables far more conversations in one fiber than what electric signals on copper cables can support. Additionally, optical fibres exhibit reduced crosstalk, meaning optical signals in one fiber do not produce undesirable effects in other adjacent fibers, ensuring clearer and more reliable communication.

Answer: expediting the local area network

Explanation: Case tools are the tools or service that helps in the software development but the automatic activity. This tool can be applied on various applications like mobile apps, E-commerce websites etc.

Analyst focus on the features like improving productivity ,low time consumption,integration of the work and making the communication better with the users ,helps in producing large amount of the documents etc.

The only factor not accepted by analyst is accomplishing the operation in the LAN(local area network) as it does not facilitates this operation because it is not software development process.

Answer: False

Explanation:

 The given statement is false because MySQL 8 basically support the JSON object since versions 5,7 and 8. The MySQL basically support the native Java script object notation (JSON) data types that is basically define by the RFC 7159 and it can efficiently access the data  from the JSON documents.

It provide many advantages like the JSON are easily detect the error in the documents and can validate automatically. It also contain optimize storage format.    

Answer:

126.375 in binary is: 1111110.011

Explanation:

In order to convert a decimal number to binary number system, the integral part is converted using the division and remainder method while the fractional part is multiplied with 2 and the integral part of answer is noted down. The fractional part is again multiplied with 2 and so on.

For 126.375

2         126

2          63 - 0

2          31 - 1

2          15 - 1

2            7 - 1

2            3 - 1

              1 - 1

So, 126 = 1111110

For 0.375

0.375 * 2 = 0.75

0.75 * 2 = 1.5

0.5 * 2 = 1.0

As we had to only find 3 digits after binary point, so

0.375 = 011

So 126.375 in binary is: 1111110.011 ..

Answer: Automated testing is testing of the software by the means of automation by using special tools .These software tools are responsible for the testing of the execution process , functioning and comparing the achieved result with actual result.

It is helps in achieving the accuracy as compared with manual testing and also also function without any human interruption.It produces the test scripts that can be re-used.

Answer:

10

Explanation:

Adding two binary numbers 0111 and 1001 results in the sum as 10000.

The equivalent hex representation for the number is obtained by splitting the result into two quads and finding their individual hex representations.

00010000

Quad 1 : 0001 = 1 (HEX)

Quad 2 : 0000 = 0 (HEX)

So the equivalent result is obtained by concatenating the two digits one after the other : 10 (in Hex)

Answer:

 The split form is the type of feature which are introduced by the Microsoft access in 2007. This feature basically give two view of the data at the similar time that is:

These two views are basically connected with the same source of the data and they are basically synchronize with the each other. From the datasheet view we can easily select the field in the record. We can also modify the data by using the functions such as add,delete and edit.