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RideAnS [48]
3 years ago
5

What type of system security malware allows for access to a computer, program, or service without authorization?

Computers and Technology
1 answer:
olasank [31]3 years ago
7 0

Answer:

"backdoor" is the correct answer for the above question

Explanation:

  • The backdoor is a type of attack which is done by the type of backdoor programs to remotely access the computers without the authorization of the user of the computer.
  • It is inserted on the computer with the help of the internet and accessed the computer without using authorization techniques.
  • The above question asked about the virus which is used to access the computer without the permission of the right use of the system. Hence the answer is backdoor which is defined above.
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4 years ago
Which of the following statements about the use of desktop publishing software is true? A. There's no visible difference between
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What is curated content?
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Answer:

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Explanation:

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6 0
2 years ago
Develop a plan to design and finally implement a set of functions using C++ that would implement the IEEE standard. Phase 1 will
AnnZ [28]

Answer:

See explaination code

Explanation:

#include<iostream>

#include<math.h>

using namespace std;

typedef union {

float number;

struct

{

// Order is important.

// Here the members of the union data structure

// use the same memory (32 bits).

// The ordering is taken

// from the LSB to the MSB.

unsigned int mantissa : 23;

unsigned int exponent : 8;

unsigned int sign : 1;

} Raw;

} MyFloat;

void printBinary(int n, int i)

{

// Prints the binary representation

// of a number n up to i-bits.

int k;

for (k = i - 1; k >= 0; k--) {

if ((n >> k) & 1)

cout << "1";

else

cout << "0";

}

}

void decToHex(int n){

// char array to store hexadecimal number

char hexaDeciNum[100];

// counter for hexadecimal number array

int i = 0;

while(n!=0)

{

// temporary variable to store remainder

int temp = 0;

// storing remainder in temp variable.

temp = n % 16;

// check if temp < 10

if(temp < 10)

{

hexaDeciNum[i] = temp + 48;

i++;

}

else

{

hexaDeciNum[i] = temp + 55;

i++;

}

n = n/16;

}

// printing hexadecimal number array in reverse order

for(int j=i-1; j>=0; j--)

cout << hexaDeciNum[j];

}

void floatBinary(float f){

long double binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;

long int integer, binaryInt = 0;

long int p = 0, rem, temp;

//separate the integer part from the input floating number

integer = (int)f;

//separate the fractional part from the input floating number

frac = f - integer;

//loop to convert integer part to binary

while (integer != 0) {

rem = integer % 2;

binaryInt = binaryInt + rem *pow(10, p);

integer = integer / 2;

p++;

}

//loop to convert fractional part to binary

while (frac != 0) {

frac = frac * 2;

temp = frac;

binaryFrac = binaryFrac + fracFractor * temp;

if (temp == 1)

frac = frac - temp;

fracFractor = fracFractor / 10;

}

cout << binaryInt + binaryFrac;

}

int findDecimal(float number){

int nfloor = number;

float nfloat = number - nfloor;

int nfloatfloor;

do {

nfloat *= 10;

nfloatfloor = nfloat;

} while (nfloat > nfloatfloor);

return nfloatfloor;

}

void first(float number){

if(number < 0)

cout << "SIGN BIT IS (1) SINCE NUMBER IS NEGATIVE" << endl;

else

cout << "SIGN BIT IS (0) SINCE NUMBER IS POSITIVE" << endl;

}

void second(float number){

cout << "INTEGER PART IN BASE-10:" << int(number) << " AND IN BINARY:";

printBinary(int(number),16);

cout << endl;

}

void third(float number){

cout << "DECIMAL PART IN BASE-10:" << findDecimal(number) << " AND IN BINARY:";

printBinary(findDecimal(number),16);

cout << endl;

}

void fourth(float number){

cout << "ENTERED NUMBER IN BASE-10:" << number << " AND IN BINARY:";

floatBinary(number);

cout << endl;

}

void fifth(MyFloat myfloat){

cout << "MANTISA IN BINARY:";

printBinary(myfloat.Raw.mantissa,32);

}

void sixth(MyFloat myfloat){

cout << "EXPONENT IN BASE-10:" << myfloat.Raw.exponent << " AND IN BINARY:";

printBinary(myfloat.Raw.exponent,8);

cout << endl;

}

void seventh(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

printBinary(myfloat.Raw.exponent,8);

cout << " | ";

printBinary(myfloat.Raw.mantissa,32);

cout << endl;

}

void eigth(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

decToHex(myfloat.Raw.exponent);

cout << " | ";

decToHex(myfloat.Raw.mantissa);

cout << endl;

}

int main(){

float number;

cout << "PLEASE ENTER A NUMBER TO DISPLAY THE IEEE 754 FLOATING POINT OPTIONS" << endl;

cin >> number;

MyFloat myfloat;

myfloat.number = number;

cout << "PLEASE CHOOSE ONE OF THE FOLLOWING OPERATIONS" << endl;

cout << " 1. DISPLAY THE SIGN BIT VALUE" << endl;

cout << " 2. DISPLAY THE INTEER PART IN BOTH BASE-10 AND CINARY FORMATS" << endl;

cout << " 3. DISPLAY THE DECIMAL PART IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 4. DISPLAY THE NUMBER ENTERED IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 5. DISPLAY THE MANTISA IN BINARY FORMATS" << endl;

cout << " 6. DISPLAY THE EXPONENT IN BORH BASEE-10 AND BINARY FORMATS" << endl;

cout << " 7. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

cout << " 8. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

int choice;

cin >> choice;

switch(choice){

case 1:first(number);

break;

case 2:second(number);

break;

case 3:third(number);

break;

case 4:fourth(number);

break;

case 5:fifth(myfloat);

break;

case 6:sixth(myfloat);

break;

case 7:seventh(myfloat);

break;

case 8:eigth(myfloat);

break;

default:cout << "ENTER VALID CHOICE" << endl;

}

}

Refer to attachment please for onscreen look.

6 0
3 years ago
Do you believe that OOP should be phased out and we should start working on some alternative(s)? Provide your answer with Yes or
o-na [289]

Answer:

I don't think so. In today's computer era, many different solution directions exist for any given problem. Where OOP used to be the doctrine of choice, now you would consider it only when the problem at hand fits an object-oriented solution.

Reason 1: When your problem can be decomposed in many different classes with each many instances, that expose complex interactions, then an OO modeling is justified. These problems typically produce messy results in other paradigms.

Reason 2: The use of OO design patterns provides a standardized approach to problems, making a solution understandable not only for the creator, but also for the maintainer of code. There are many OO design patterns.

5 0
3 years ago
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