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Paladinen [302]
2 years ago
14

Electronic evidence on computer storage media that is not visible to the average user is called​ ________.

Computers and Technology
1 answer:
Nookie1986 [14]2 years ago
5 0

Answer:

Electronic evidence on computer storage media that is not visible to the average user is called​ ambient data.

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The cost of a CLEP exam is about__<br><br> A$100 <br> B $50 <br> C$80<br> D$170
Elanso [62]

Your answer is C or $80

When averaging the costs of credit hours at public and private community colleges and four-year institutions, the median cost for a three-hour class in 2017 was $1,782. Conversely, each CLEP exam costs $85.

5 0
3 years ago
Read 2 more answers
class llist { private: record * start; char filename[16]; int readfile(); int writefile(); record * reverse(record * ); void cle
Sophie [7]

Answer:

See Explaination

Explanation:

#include <cstdlib>

#include <iostream>

#include <iomanip>

#include <fstream>

#include <string>

#include <sstream>

using namespace std;

struct Person

{

string name;

string addr;

string phone;

string id;

};

const int MAX_SIZE = 8;

void addData(Person list[], int& size);

void dispData(const Person list[], int size);

void delData (const Person list[], int size);

void saveFile(const Person list[], int size);

void openFile(Person list[], int& size);

char getMenuResponse();

int main(int argc, char *argv[])

{

Person contactList[MAX_SIZE];

int numOfRecs = 0;

bool run = true;

do

{

cout << "Address Book - " << numOfRecs << " Number of Contacts" << endl;

switch ( getMenuResponse() )

{

case 'A': addData(contactList, numOfRecs); break;

case 'D': dispData(contactList, numOfRecs); break;

case 'T': delData(contactList, numOfRecs); break;

case 'O': openFile(contactList, numOfRecs); break;

case 'S': saveFile(contactList, numOfRecs); break;

case 'Q': run = false; break;

default : cout << "That is NOT a valid choice" << endl;

}

} while (run);

cout << endl << "Program Terminated" << endl;

return EXIT_SUCCESS;

}

void addData(Person list[], int& size)

{

Person tmp;

char response;

char str[256];

if (size < MAX_SIZE) {

system("cls");

cout << "Enter Contact Information" << endl << endl;

cout << "Name: ";

cin.getline(str, 256, '\n');

tmp.name = str;

cout << endl;

cout << "Address: ";

cin.getline(str, 256, '\n');

tmp.addr = str;

cout << endl;

cout << "Phone Number: ";

cin.getline(str, 256, '\n');

tmp.phone = str;

cout << endl;

cout << "E-mail id Address: ";

cin.getline(str, 256, '\n');

tmp.id = str;

cout << endl;

cout << "Add Contact to Address Book? (y/n) ";

cin >> response;

if (toupper(response) == 'Y')

list[size++] = tmp;

} else {

cout << "Sorry, Address Book is currently full." << endl;

system("pause");

}

system("cls");

}

void dispData(const Person list[], int size)

{

system("cls");

if(size < 1)

{

cout << "Nothing to display" << endl;

} else {

cout << "Contacts :" << endl << endl;

cout << fixed << setprecision(2);

cout << "Contact Name Address Phone No. ID" << endl;

cout << left;

for (int i = 0; i < size; i++)

{

cout << setw(10) << list[i].name << right

<< setw(10) << list[i].addr<<right

<< setw(10) << list[i].phone<<left

<< setw(15) << list[i].id<<left<<endl;

}

cout << right << setw(3) << size;

cout << " Contacts"<< endl;

}

system("PAUSE");

system("cls");

}

void delData(const Person list[],int size) {

vector<Person> :: iterator vItr = Contact.begin();

while (vItr != Contact.end() )

{

if (vItr->Name == Name)

{

vItr = Contact.erase (vItr);

break;

}

else

vItr++;

}

void saveFile(const Person list[], int size) {

string fileName;

ofstream outfi;

cout<<"Enter file name: ";

getline(cin,fileName);

outfi.open(fileName.c_str());

if (!outfi.fail()) {

system("cls");

cout << "Saving Address Book to the disc ";

for(int i = 0; i < size; i++) {

outfi << list[i].name << ';'

<< list[i].addr<< ';';

if (i < size-1) outfi << endl;

}

cout << endl << size << " Address Book in the disc." << endl;

outfi.close();

system("PAUSE");

system("cls");

}

else {

cout << "ERROR: problem with file" << endl;

system("PAUSE");

system("cls");

}

}

void openFile(Person list[], int& size)

{

ifstream infi("AddressBook.txt");

string str;

stringstream strstrm;

if (!infi.fail()) {

system("cls");

cout << "Reading Address Book from the disc ";

size = 0;

while(!infi.eof() && size < MAX_SIZE)

{

getline(infi, str, ';');

list[size].name = str;

getline(infi, str, ';');

strstrm.str(""); strstrm.clear();

strstrm << str;

strstrm >> list[size].addr;

}

cout << endl << size << " read contacts from the disc." << endl;

system("PAUSE");

system("cls");

}

else {

cout << "ERROR :file not accepted" << endl;

system("PAUSE");

system("cls");

}

}

char getMenuResponse()

{

char response;

cout << endl << "take chioce" << endl

<< "(A)dd contact, (D)isplay contact0, (O)pen File, (S)ave File, (Q)uit" << endl

<< "> ";

cin >> response;

cin.ignore(256, '\n');

return toupper(response);

}

8 0
3 years ago
The ______ view connects perceptual capabilities to information available in the world of the perceiver. Multiple choice questio
jekas [21]

Answer:

Ecological

Explanation:

Please mark brainliest

4 0
2 years ago
You accidentally moved your task bar from the bottom of the screen to the left side. You would like to
Ratling [72]

Answer:

by pressing yes

Explanation:

ik i am a tech god

6 0
3 years ago
Read 2 more answers
Part 3:<br> Please answer
Levart [38]

Answer:

a) 4 processes

b) 2 resources

c) R1: 2 instances

   R2: 2 instances

d) R2

e) R1

f) R1

g) No resource

h)  R1

i) R2

j) R2

k) No

l)No deadlock

Explanation:

You need to know that the resources that are required by the processes for completion are shown by the request edge and the resources allocated are shown by the allocation edge. And thus, we can find what resources are allocated to the process, and required for the completion accordingly. And here again, the cycle is created, and hence deadlock may or may not occur. However, we see that resources have multiple instances and get freed on time. And hence, deadlock does not occur.

Like,

P4 uses R2 and free one instance of R2.

P3 then uses one instance of R2 and free R2.

P3 then uses one instance of R3 and free R3.

P1 uses one instance of R1 and free R1 one instance

P1 then uses R2 and free R2.

P2 uses R1 and free R1.

Hence, all the processes are complete and deadlock does not occur.

3 0
3 years ago
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