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2 years ago

Which of these is NOT a type of spoofing? sending an e-mail that says it is from a certain bank but it is not

Computers and Technology

1 answer:

ArbitrLikvidat [17]2 years ago

3 0

The option that is not a type of spoofing is creating a fake website for a well-known business.

<h3>What is email spoofing?</h3>

Email spoofing is known to be a method that is often used in spam and other forms of phishing attacks. This is often done to trick computer users to see that a message originate from a person or firm that they know.

Conclusively, The option that is not a type of spoofing is creating a fake website for a well-known business as it is not what spoofing is all about.

Learn more about spoofing from

brainly.com/question/23021587

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What are foundations of any game systems that control what the players can or cannot do in the game, as well as the penalties, r

san4es73 [151]

Answer:

A. Rules

Explanation:

Rules can determine what can be done and what cannot be done... giving it restrictions, or RULES in a video game among-st a player

4 0

3 years ago

Given a list ` = (x1 x2 . . . xn−1 xn), we define two rotations of the list: The left-rotation is (x2 . . . xn−1 xn x1) and, lik

aleksandrvk [35]

Explanation:

The below code has been written in C language

void rotateright(int list[], int n)

{

int x = list[n-1]

int i;

for (i = n-1; i > 0; i--)

list[i] = list[i-1];

list[0] = x;

}

void rotateleft(int list[], int n)

{

int x = list[0]

int i;

for (i = 1; i < n-1 ; i++)

list[i] = list[i+1];

list[n-1] = x;

}

int main()

{

int list[] = {x1, x2, x3, ... x(n-1),xn}

int i;

int n = sizeof(list);

rotateright(list, n);

rotateleft(list, n);

return 0;

}

3 0

3 years ago

A musician has recorded some initial ideas for new songs which she wishes to share with her bandmates. As these are initial idea

Ugo [173]

Answer:

1.save the audio using a low sampling rate

2.save the audio using a low bit depth

Explanation:

1. if the quality of the audio is low then the size of the audio will also be low and which will make the size of the data to be less and also easier to download

3 0

3 years ago

Write a program that calculates the average rainfall for three months. The program should ask the user to enter the name of each

Eduardwww [97]

Answer:

import java.util.Scanner;

public class ANot {

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

System.out.println("Enter the month name:");

String month1 = input.next();

System.out.println("Enter the amount of rain for "+month1);

double month1Rain = input.nextDouble();

System.out.println("Enter the second month: ");

String month2 = input.next();

System.out.println("Enter the amount of rain for "+month2);

double month2Rain = input.nextDouble();

System.out.println("Enter the Third month: ");

String month3 = input.next();

System.out.println("Enter the amount of rain for "+month3);

double month3Rain = input.nextDouble();

double averageRainfall = (month1Rain+month2Rain+month3Rain)/3;

System.out.printf("Average Rainfall is: %,.2f ", averageRainfall);

}

Explanation:

This is written in Java Programming language

Import the scanner class to prompt user for the inputs (Months and amount of rainfall for each month)
Find the average of the three rainfall amount by adding up and dividing by 3
Output the average. Pay particular attention to the use of java's printf in order to print the average to 2 decimal places

8 0

3 years ago

Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera

zalisa [80]

Answer:

Explanation:

#include<iostream>

#include<ctime>

#include<bits/stdc++.h>

using namespace std;

double calculate(double arr[], int l)

{

double avg=0.0;

int x;

for(x=0;x<l;x++)

{

avg+=arr[x];

}

avg/=l;

return avg;

}

int biggest(int arr[], int n)

{

int x,idx,big=-1;

for(x=0;x<n;x++)

{

if(arr[x]>big)

{

big=arr[x];

idx=x;

}

return idx;

}

int main()

{

vector<pair<int,double> >result;

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

int choice;

cin>>choice;

while(choice!=2)

{

int n,m;

cout<<"Enter N"<<endl;

cin>>n;

cout<<"Enter M"<<endl;

cin>>m;

int c=m;

double running_time[c];

while(c>0)

{

int arr[n];

int x;

for(x=0;x<n;x++)

{

arr[x] = rand();

}

clock_t start = clock();

int pos = biggest(arr,n);

clock_t t_end = clock();

c--;

running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;

}

double avg_running_time = calculate(running_time,m);

result.push_back(make_pair(n,avg_running_time));

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

cin>>choice;

}

for(int x=0;x<result.size();x++)

{

cout<<result[x].first<<" "<<result[x].second<<endl;

}

8 0

3 years ago