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3 years ago

Why are many otherwise safety-conscious people victims of Internet crime?

Computers and Technology

1 answer:

kkurt [141]3 years ago

5 0

Because safety-conscious people don't install virus protectors. Hope this helps.

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What is the output of the following C++ program?

KIM [24]

Answer:

Output: 15 11/16 inches

Explanation:

#include <iostream>

using namespace std;

////////////////////////////////////////////////////////////////////////

class InchSize {

public:

InchSize(int wholeInches = 0, int sixteenths = 0);

void Print() const;

InchSize operator+(InchSize rhs);

private:

int inches;

int sixteenths;

};

InchSize InchSize::operator+(InchSize rhs) {

InchSize totalSize;

totalSize.inches = inches + rhs.inches;

totalSize.sixteenths = sixteenths + rhs.sixteenths;

// If sixteenths is greater than an inch, carry 1 to inches.

if (totalSize.sixteenths >= 16) {

totalSize.inches += 1;

totalSize.sixteenths -= 16;

}

return totalSize;

}

InchSize::InchSize(int wholeInches, int sixteenthsOfInch) {

inches = wholeInches;

sixteenths = sixteenthsOfInch;

}

void InchSize::Print() const {

cout<<inches<<" "<<sixteenths<<"/16 inches"<<endl;

}

////////////////////////////////////////////////////////////////////////

int main() {

InchSize size1(5, 13);

InchSize size2(9, 14);

InchSize sumSize;

sumSize = size1 + size2;

sumSize.Print();

return 0;

}

////////////////////////////////////////////////////////////////////////

sumSize variable was printed in the end. Print() prints the calling object's inches and sixteenths variables' values.

sumSize's inches is equal to size1 plus size2's inches.

Because the sum of sixteenths was greater than 16, sumSize's sixteenth was decreased by 1 and inches was increased by 1.

3 0

3 years ago

You can place an insertion point by clicking in the field or by pressing ____.

Slav-nsk [51]

You can place an insertion point by clicking in the field or by clicking F2 keyboard shortcut. Insertion point is usually characterized by a blinking vertical line that allows you to insert a next character that you wanted.

4 0

3 years ago

You can use the ____ utility to zero in on the service or other program that is slowing down startup. 1. gpupdate 2. MSconfig 3.

Step2247 [10]

I'd say MSconfig, you can use that and view the 'processes' tab, but it will make you open up the task manager.

3 0

3 years ago

Write c++ program to find maximum number for three variables using statement ?

pantera1 [17]

Answer:

#include<iostream>

using namespace std;

int main(){

int n1, n2, n3;

cout<<"Enter any three numbers: ";

cin>>n1>>n2>>n3;

if(n1>=n2 && n1>=n3){

cout<<n1<<" is the maximum";}

else if(n2>=n1 && n2>=n3){

cout<<n2<<" is the maximum";}

else{

cout<<n3<<" is the maximum";}

return 0;

}

Explanation:

The program is written in C++ and to write this program, I assumed the three variables are integers. You can change from integer to double or float, if you wish.

This line declares n1, n2 and n3 as integers

int n1, n2, n3;

This line prompts user for three numbers

cout<<"Enter any three numbers: ";

This line gets user input for the three numbers

cin>>n1>>n2>>n3;

This if condition checks if n1 is the maximum and prints n1 as the maximum, if true

if(n1>=n2 && n1>=n3){

cout<<n1<<" is the maximum";}

This else if condition checks if n2 is the maximum and prints n2 as the maximum, if true

else if(n2>=n1 && n2>=n3){

cout<<n2<<" is the maximum";}

If the above conditions are false, then n3 is the maximum and this condition prints n3 as the maximum

else{

cout<<n3<<" is the maximum";}

return 0;

3 0

3 years ago

2.Consider the following algorithm and A is a 2-D array of size ???? × ????: int any_equal(int n, int A[][]) { int i, j, k, m; f

Elis [28]

Answer:

(a) What is the best case time complexity of the algorithm (assuming n > 1)?

Answer: O(1)

(b) What is the worst case time complexity of the algorithm?

Answer: O(n^4)

Explanation:

(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1) .

(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).

5 0

3 years ago