If you were required to give a speech identifying the risks of using computers and digital devices, which group of items would you include?

Answer:

Explanation:

public class Team {

   private String teamName;

   private int teamWins;

   private int teamLosses;

   public String getTeamName() {

       return teamName;

   }

   public void setTeamName(String teamName) {

       this.teamName = teamName;

   }

   public int getTeamWins() {

       return teamWins;

   }

   public void setTeamWins(int teamWins) {

       this.teamWins = teamWins;

   }

   public int getTeamLosses() {

       return teamLosses;

   }

   public void setTeamLosses(int teamLosses) {

       this.teamLosses = teamLosses;

   }

   public double getWinPercentage() {

       return teamWins / (double) (teamWins + teamLosses);

   }

}

Following are the Java program to define the Team class and calculate its  value:

class Team //defining the class Team

{

   private String teamName;//defining String variable

   private int teamWins, teamLosses;//defining integer variable

   //defining the set method to set value the input value

   public void setTeamName(String teamName)//defining setTeamName method that takes one String parameter

   {

       this.teamName = teamName;//using this keyword that sets value in teamName

   }

   public void setTeamWins(int teamWins) //defining setTeamWins method that takes one integer parameter

   {

       this.teamWins = teamWins;//using this keyword that sets value in teamWins

   }

   public void setTeamLosses(int teamLosses)//defining setTeamLosses method that takes one integer parameter

   {

       this.teamLosses = teamLosses;//using this keyword that sets value in teamLosses

   }

   //defining the get method that returns the input value

   public String getTeamName() //defining getTeamName method

   {

       return teamName;//return teamName value

   }

   public int getTeamWins()  //defining getTeamWins method

   {

       return teamWins;//return teamWins value

   }

   public int getTeamLosses() //defining getTeamLosses method

   {

       return teamLosses;//return teamLosses value

   }

   public double getWinPercentage()//defining getWinPercentage method

   {              

       return ((teamWins * 1.0) / (teamWins + teamLosses));//using the return keyword that returns percentage value

   }      

}

Please find the complete code in the attached file and its output file in the attached file.

Class definition:

Find out more about the Class here:

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Answer:

cyberspace

Explanation:

It is the notional environment in which communication over computer networks occurs.

It is imperative that we safeguard this domain known as cyberspace.

The meaning of the cyberspace is the climate of the Internet.

There is large network of computers connected together in offices or organizations to explore the internet. The cyberspace has formed the first important needed thing in this century without which nothing can be done in personal space or in any companies.

The U.S. continues to become more dependent on the global domain within the information environment consisting of the interdependent network of information technology infrastructures, including the Internet, telecommunications networks, computer systems, and embedded processors and controllers.

So, we safeguard this domain known a cyberspace.

Learn more about cyberspace.

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Answer:

The pseudocode is as given below in the explanation while the flow diagram is attached herewith

Explanation:

The pseudocode is as follows

input cust_name, num_texts  

Set Main_bill=5 $;

if the num_texts is less than or equal to 60

Excess_Charge=0;

If the num_texts is greater than 60 and less than or equal to 200

num_text60200 =num_texts-60;

Excess_Charge=0.1*(num_text60200);

else If the num_texts is greater than 60

num_texts200 =num_texts-200;

Excess_Charge=0.25*(num_texts200)+0.1*(200-60);

Display cust_Name

Total_Bill=Main_bill+Excess_Charge;

Total_Bill_after_Tax=Total_Bill*1.12;

Answer:

Full back up

It creates complete copy of the source of data. This feature made it to the best  back  up with good speed of recovery of the data and  simplicity.However, because it  is needed for backing up  of  a lot of data, it is time consuming process,Besides it also  disturbs the routine application of  the technologist infrastructure. In addition large capacity storage is needed  for the  storage of  the large volume of back up with each new creation.

Incremental back up: back up work at high speed thus saves time, Little storage capacity is needed. This type of back up can be run as  many time as wished.

However, the restoration of data takes time, due to the time spent for restoration of initial back up and the incremental.

Differential Back up has very fast back up operation, because it only requires two pieces of back up.This is however slower compare to the Incremental.

The selective back up enable selection of certain type of file or folders for back up.

it can only be used to back up folders only when important data  are needed only.Thus the  storage capacity is low.

CPD eliminates back up window and reduces the recovery point objective.However, it has issue of compatibility with the resources to back up.,it is also expensive because its solutions are disk based.

Cloud based is affordable and save cost,It is easier to access. it is efficient,it can be access remotely for convenience.

However, Internet sources and some bandwidth must be available for access., always depends on third party providers,There is also problems encountered when switching providers.

Explanation:

Answer:

Engineering and Science

Explanation:

Answer:

In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.

Explanation:

a. main calls fun1; fun1 calls fun2; fun2 calls fun3

Solution:

b. main calls fun1; fun1 calls fun3

Solution:

c. main calls fun2; fun2 calls fun3; fun3 calls fun1

Solution:

Here variables c, d and e of fun2 are not visible .

d. main calls fun1; fun1 calls fun3; fun3 calls fun2

Solution:

The specification should be developed to help your mother buy her hardware and software on her own is hardware.

Hardware refers back to the computer's tangible additives or shipping structures that shop and run the written commands furnished through the software program. The software program is the intangible a part of the tool that shall we the consumer have interaction with the hardware and command it to carry out unique tasks.

Read more about the hardware :

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Answer: seen below

Explanation:

There are various solution or ways to go about this actually. In the end, curtailing this is what matters in the end.

Below i have briefly highlighted a few precise way some of this problems can be solved, other options are open for deliberation.

A. Digital encoding of information with the calculation being changed occasionally, particularly after the frameworks advisors have finished their employments, and the framework is being used.

Which can also mean; The money could be withdrawn form the ATM. preventive Control is password to be changed periodically and regularly.

B. Noise on the line might be causing line blunders, which can bring about information misfortune. Reverberation checks and equality checks can assist with recognizing and right such mistakes.

C. If information is being lost, reverberation checks and equality checks ought to likewise help; notwithstanding, the issue might be that an intruder is blocking messages and altering them. Message arrangement numbering will assist with deciding whether messages are being lost, and in the event that they are maybe a solicitation reaction strategy ought to be executed that makes it hard for gatecrashers to dodge.

D. In the event that messages are being postponed, a significant client request or other data could be missed. As in thing c, message arrangement numbering and solicitation reaction methods ought to be utilized.

E. Messages modified by intruders can have an extremely negative effect on client provider relations if orders are being changed. For this situation, information encryption is important to keep the gatecrasher from perusing and changing the information. Additionally, a message succession numbering strategy is important to ensure the message isn't erased.

cheers i hope this was helpful !!

Answer:

The correct answer is option (c) 4950 cipher keys

Explanation:

Solution

Given that:

From the given question we have that if there are 100 computers in a LAN, then each computer should have how many keys.

Now,

The number of computers available = 100

The number of keys used in  symmetric key cipher for N parties is given as follows:

= N(N-1)/2

= 100 * (100 -1)/2

= 50 * 99

= 4950 cipher keys

Find the given attachments

Answer:

Sequencing

Explanation:

Answer:

SEQUENCING

Explanation:

Answer:

Given: K(x,y) denotes statement:

"x knows y"

D denote domain of all people.

Explanation:

1. Everybody knows somebody.

Solution:

∀x∈D ∃y∈D : K(x, y)

∀ means for all. Here it is used for Everybody.

∃ means there exists some. Here it represents Somebody.

∈ means belongs to . Both x and y belongs to the domain D of all people.

2. There is somebody that no one knows.

Solution:

∀x∈D ∃y∈D : ¬K(x, y)

∀ means for all. ∃ means there exists some. ∈ means belongs to both x and y belongs to the domain D of all people.¬ this is negation sign which means not K(x,y). So the negation of everybody knows somebody can be expressed as there is somebody that no one knows.

3. There is no one who knows everybody

Solution:

This can be represented in both the ways below.

∀ y∈D ∃ x∈D : K(x, y)

∀ means for all. ∃ means there exists some. ∈ means belongs to both x and y belongs to the domain D of all people.

∀x∈D ∀y∈D : ¬K(x, y)

∀ means for all. The negation shows that there is no one who knows everybody.

Explanation:

A transaction is a very small system unit that can contain many low-level tasks. A transaction in a database system should maintain, Atomicity, Consistency, Isolation, and Durability − these are commonly known as ACID properties − in order to ensure accuracy, completeness, and integrity of the data.

An example of a simple transaction is as below,  Suppose a bank employee transfers Rs 500 from A's account to B's account.

A’s Account

Open_Account(A)

Old_Balance = A.balance

New_Balance = Old_Balance - 500

A.balance = New_Balance

Close_Account(A)

B’s Account

Open_Account(B)

Old_Balance = B.balance

New_Balance = Old_Balance + 500

B.balance = New_Balance

Close_Account(B)

Answer:

Explanation:

The objective here is to determine if the programmer can simply increase the upperbound to 80,000.

Of course Yes, The programmer can simply increase the delay by doubling the upperbound by 80000. The representation can be illustrated as:

( int : i = 0;  i <  40,000; i ++ )

{

  // delay code

}

Which can be modified as:

( int : i = 0;  i <  80,000; i ++ )

{

  // delay code

}

b)  If the code contains a delay loop and we noticed that no delay is being created at run-time. What should we suspect during debugging?

Assuming there is no delay being created at the run-time,

The code is illustrated as:

For ( int : i = 0 ; i < 0 ; i ++ )

{

  // delay code which wont

  //execute since code delay is zero

}

we ought to check whether the loop is being satisfied or not.  At the Initial value of loop variable, is there any break or exit statement is being executed in between loop. Thus, the  aforementioned delay loop wont be executed since the loop wont be executed for any value of i.

Answer:

Sorted Array A { } = { 1, 4, 23, 32, 34, 34, 67, 78, 89, 100 }

Explanation:

Binary tree is drawn given that the binary tree do not follow both minimum heap and maximum heap property, therefore, it is not a heap.

See attached picture.

Answer:

Replace /* Your solution goes here */  with the following lines of code

for(i = 0;i<SCORES_SIZE-1;i++)

{

bonusScores[i]+=bonusScores[i+1];

}

Explanation:

The above iteration starts from the index element (element at 0) and stops at the second to the last element (last - 1).

Using an iterative variable, i

It adds the current element (element at i) with the next element; element at i + 1.

The full code becomes

#include<iostream>

using namespace std;

int main(void) {

const int SCORES_SIZE = 4;

int bonusScores[SCORES_SIZE];

int i = 0;

bonusScores[0] = 10;

bonusScores[1] = 20;

bonusScores[2] = 30;

bonusScores[3] = 40;

for(i = 0;i<SCORES_SIZE-1;i++)

{

bonusScores[i]+=bonusScores[i+1];

}

/* Your solution goes here */

for (i = 0; i < SCORES_SIZE; ++i) {

printf("%d ", bonusScores[i]);

}

printf("\n");

return 0;

}

See attachment for .cpp file

Answer:int main() {

const int SCORES_SIZE = 4;

int bonusScores[SCORES_SIZE];

int i;

for (i = 0; i < SCORES_SIZE; ++i) {

cin >> bonusScores[i];

}

for (i = 0; i < SCORES_SIZE-1; ++i){

bonusScores[i] += bonusScores[i+1];

}

for (i = 0; i < SCORES_SIZE; ++i) {

cout << bonusScores[i] << " ";

}

cout << endl;

return 0;

}

Explanation: SCORES_SIZE -1 will prevent the for loop from going past the last value in the array. bonusScores[i] += will add the value of bonusScores[i+1] to the original bonusScores[i].

for example, i = 1; 1 < SCORES_SIZE - 1 ; bonusScores[1] += bonusScores[1+1} becomes{ bonusScores[1] + bonusScores{2];

Answer:

None

Explanation:

These are all stereotypes. Sure, there are definitely some people who fit their stereotypes, but not all. It's just a generalization at the end of the day. I can't really agree with any of them due to the fact that it's all stereotyping.

Perhaps you feel differently, and believe that some of these example are true. I can't though, sorry. Hope this take helps.

Answer:Full-interruption--A

Explanation:  The Full-interruption is one of the major steps for  a Disaster Recovery Plan, DRP which ensures businesses are not disrupted by saving valuable resources during a disaster like a data breach from fire or flood.

Although expensive and very risky especially in its simulation of a disruption, this thorough  plan ensures that when a disaster occurs, the  operations are shut down at the primary site and are transferred to the recovery site allowing Individuals follow every procedure, ranging from the interruption of service to the restoration of data from backups, also with the notification of appropriate individuals.

Answer and Explanation:

Answer:

CASE tools reduce the need for end user participation in systems development.

Explanation:

CASE is an acronym for Computer-aided Software Engineering and it comprises of software application tools that provide users with automated assistance for Software Development Life Cycle (planning, analysing, designing, testing, implementation and maintenance). The CASE tools helps software developers in reducing or cutting down of the cost and time of software development, as well as the enhancement of the software quality.

Some other benefits of using the CASE tools are;

- CASE tools provide automated graphics facilities for producing charts.

- CASE tools have capabilities for validating design diagrams and specifications.

- CASE tools support collaboration among team members.

- CASE tools facilitate the creation of clear documentation.

- CASE tools checks for consistency, syntax errors and completeness.

The CASE tools can be grouped as, requirement and structure analysis, software design, test-case and code generation, reverse engineering, and document production tools.

Examples of CASE tools are flowchart maker, visible analyst (VA), dreamweaver, net-beans, microsoft visio, adobe illustrator and photoshop etc.

Answer: you want to list all LC structers

Explanation:

Answer:

LAN/WAN domain

Explanation:

This is the boundary between trusted adn untrusted zones. All SLA's, managed security services, their monitoring and reporting is done by this domain.

Answer:

Option (d) is correct

Explanation:

Previously saved capture files can be read using Wireshark. For this, select the File then open the menu. Then Wireshark will pop up the “File Open” dialog box.

Wireshark capture files, like the DemoCapturepcap file found in this lab, have a .pcapng extension, which stands for packet capture, next generation.

Answer:

maybe threat?

Explanation:

Answer:

The program written in C language is as follows

#include<stdio.h>

int main()

{

//Declare digit

int digit;

//Prompt user for input

printf("Enter any integer: [10 - 100]: ");

scanf("%d", &digit);

//Check if digit is within range 10 to 100

while(digit<10 || digit >100)

{

printf("Enter any integer: [10 - 100]: ");

scanf("%d", &digit);

}

//Initialize counter to 0

int counter = 0;

for(int i=digit;i>0;i--)

{

 printf("%3d", i); //Print individual digit

 counter++;

 if(counter == 10) //Check if printed digit is up to 10

 {

  printf("\n"); //If yes, print a new line

  counter=0; //And reset counter to 0

 }

}

}

Explanation:

int digit; ->This line declares digit as type int

printf("Enter any integer: [10 - 100]: "); -> This line prompts user for input

scanf("%d", &digit);-> The input us saved in digit

while(digit<10 || digit >100) {

printf("Enter any integer: [10 - 100]: ");

scanf("%d", &digit); }

->The above lines checks if input number is between 10 and 100

int counter = 0; -> Declare and set a counter variable to 0

for(int i=digit;i>0;i--){ -> Iterate from user input to 0

printf("%3d", i); -> This line prints individual digits with 3 line spacing

counter++; -> This line increments counter by 1

if(counter == 10){-> This line checks if printed digit is up to 10

printf("\n"); -> If yes, a new line is printed

counter=0;} -> Reset counter to 0

} - > End of iteration

Answer:

b. False

Explanation:

The system of fraud is called 4-1-9 scam. And yes, victims are always promised large sum of money to secretly help in moving a large sum of money.

Answer:

The csharp program is as follows.

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

using System.Windows.Forms;  

namespace WindowsFormsApplication1

{

   public partial class Form1 : Form

   {

       public Form1()

       {

           InitializeComponent();

       }

private void button1_Click(object sender, EventArgs e)

       {

           string word = textBox1.Text;

               string  ch = word.Substring(0, 1);

               string str = word.Substring(1, word.Length-1);

               string s = str.Insert(str.Length, ch);

               textBox2.Text = s.Insert(s.Length, "ay");            

       }

     private void button3_Click(object sender, EventArgs e)

       {

           Close();

       }

       private void button2_Click(object sender, EventArgs e)

       {

           textBox1.Text = "";

           textBox2.Text = "";

       }

   }

}

Explanation:

1. A string variable to hold the user input is declared and initialized accordingly. The user inputted string is taken from textbox1.

string word = textBox1.Text;

2. A string variable to hold the first character of the user inputted string is declared and initialized.

string  ch = word.Substring(0, 1);

3. A string variable to hold the user inputted string without the first character is declared and initialized accordingly.

string str = word.Substring(1, word.Length-1);

4. A string variable to hold the substring from step 3 along with the inserted characters at the end, is declared and initialized accordingly.

string s = str.Insert(str.Length, ch);

5. The final string is assigned to the textbox 2, which is the PigLatin conversion of the user inputted string.

textBox2.Text = s.Insert(s.Length, "ay");

6. All the above take place when the user clicks Convert to PigLatin button.

7. Two additional buttons, clear and exit are also included in the form.  

8. When the user clicks clear button, both the textboxes are initialized to empty string thus clearing both the textboxes.

 textBox1.Text = "";

           textBox2.Text = "";

9. When the user clicks the exit button, the application closes using the Close() method.

10. The program is done in Visual Studio.

11. The output of the program is attached.

12. The program can be tested for any type of string and any length of the string.

Answer: provided in the explanation section

Explanation:

The question says:

Write a short assembly language program in either our 8088 SCO DOSBox or 80386+ MASM Visual Studio 2017 environment that demonstrates data storage and retrieval from memory. As an example consider some value which is either 16 or 32 bits, after instantiating as an immediate value transfer to and retrieve from memory using assembly language instructions.

The Answer:

multi-segment executable file template. data segment ; add your data here! pkey db "press any key...$" ends stack segment dw 128 dup(0) ends code segment start: ; set segment registers: mov ax, data mov ds, ax mov es, ax ; add your code here mov cx,4 input: mov ah,1 int 21h push ax loop input mov dx,13d mov ah,2 int 21h mov dx,10d mov ah,2 int 21h mov cx,4 output: pop bx mov dl,bl mov ah,2 int 21h loop output exit: lea dx, pkey mov ah, 9 int 21h ; output string at ds:dx ; wait for any key....

mov ah, 1 int 21h mov ax, 4c00h ; exit to operating system. int 21h ends end start ;

set entry point and stop the assembler.

Cheers I hope this helps!!!

Answer:

Explanation:

import java.io.*;

class Triangle

{

  private double side1, side2, side3; // the length of the sides of

                                       // the triangle.

  //---------------------------------------------------------------

  // constructor

  //

  // input : the length of the three sides of the triangle.

  //---------------------------------------------------------------

  public Triangle(double side1, double side2, double side3)

  {

      this.side1 = side1;

      this.side2 = side2;

      this.side3 = side3;

  }

  //---------------------------------------------------------------

  // isRight

  //

  // returns : true if and only if this triangle is a right triangle.

  //---------------------------------------------------------------

  boolean isRight()

  {

      double square1 = side1*side1;

      double square2 = side2*side2;

      double square3 = side3*side3;

      if ((square1 == square2 + square3) ||

          (square2 == square1 + square3) ||

          (square3 == square1 + square2))

          return true;

      else

          return false;

  }

  // isValid

  // returns : true if and only if this triangle is a valid triangle.

  boolean isValid()

  {

      if ((side1 + side2 < side3) ||

          (side1 + side3 < side2) ||

          (side2 + side3 < side1))

          return false;

      else

          return true;

  }

  // isEquilateral

  //

  // returns : true if and only if all three sides of this triangle

  // are of the same length.

  boolean isEquilateral()

  {

      if (side1 == side2 && side2 == side3)

          return true;

      else

          return false;

  }

  // isIsosceles

  //

  // returns : true if and only if exactly two sides of this triangle

  // has the same length.

  boolean isIsosceles()

  {

      if ((side1 == side2 && side2 != side3) ||

          (side1 == side3 && side2 != side3) ||

          (side2 == side3 && side1 != side3))

          return true;

      else

          return false;

  }

  // isIsosceles

  // returns : true if and only if exactly no two sides of this

  // triangle has the same length.

  boolean isScalene()

  {

      if (side1 == side2 || side2 == side3 || side1 == side3)

          return false;

      else

          return true;

  }

}

//-------------------------------------------------------------------

// class Application

//

// This class is the main class of this application. It prompts

// the user for input to construct a triangle, then prints out

// the special properties of the triangle.

//-------------------------------------------------------------------

public class Application

{

  //---------------------------------------------------------------

  // getInput

  //

  // input : stdin - BufferedReader to read input from

  //         msg - message to prompt the user with

  // returns : a double value input by user, guranteed to be

  //           greater than zero.

  //---------------------------------------------------------------

  private static double getInput(BufferedReader stdin, String msg)

      throws IOException

  {

      System.out.print(msg);

      double input = Double.valueOf(stdin.readLine()).doubleValue();

      while (input <= 0) {

          System.out.println("ERROR : length of the side of triangle must " +

              "be a positive number.");

          System.out.print(msg);

          input = Double.valueOf(stdin.readLine()).doubleValue();

      }

      return input;

  }

  //---------------------------------------------------------------

  // printProperties

  //

  // input : triangle - a Triangle object

  // print out the properties of this triangle.

  //---------------------------------------------------------------

  private static void printProperties(Triangle triangle)

  {

      // We first check if this is a valid triangle. If not

      // we simply returns.

      if (!triangle.isValid()) {

          System.out.println("This is not a valid triangle.");

          return;

      }

      // Check for right/equilateral/isosceles/scalene triangles

      // Note that a triangle can be both right triangle and isosceles

      // or both right triangle and scalene.

      if (triangle.isRight())

          System.out.println("This is a right triangle.");

      if (triangle.isEquilateral())

          System.out.println("This is an equilateral triangle.");

      else if (triangle.isIsosceles())

          System.out.println("This is an isosceles triangle.");

      else

          // we do not need to call isScalene here because a triangle

          // is either equilateral/isosceles or scalene.

          System.out.println("This is an scalene triangle.");

  }

  // main

  // Get the length of the sides of a triangle from user, then

  // print out the properties of the triangle.

  public static void main(String args[]) throws IOException

  {

      BufferedReader stdin = new BufferedReader

          (new InputStreamReader(System.in));

      double side1 = getInput(stdin,

          "What is the length of the first side of your triangle? ");

      double side2 = getInput(stdin,

          "What is the length of the second side of your triangle? ");

      double side3 = getInput(stdin,

          "What is the length of the third side of your triangle? ");

      System.out.print("Pondering...\n");

      printProperties(new Triangle(side1, side2, side3));

  }

}

Answer:

Explanation:

MIPS program which increments from 0 to 15 and display results in Decimal on the console

In this program the user defined procedures print_int and print_eot were used to print the integer values and new line characters(\n) respectively.the mechanisam of the loop is explaine in the comment section of the program.

     addi $s0, $0, 0

     addi $s1, $0, 15

print_int:

              li $v0, 1                # system call to print integer

              syscall                

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s1<$s0

beq $s0, $0, end

end:

MIPS progam to increment from 0 to 15 and display results in Hexadecimal on the console

this program is slightly differed from the previous program in this program the system call issued in print_int is implemented with a system call that prints numbers in hex.

addi $s0, $0, 15

     addi $s1, $0, 0

print_int:

   li      $v0,34                  # syscall number for "print hex"

   syscall                         # issue the syscall

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s0>$s1

beq $s0, $0, end

end: