<span>Operating System is the software that gives the computer instructions on how to run applications. It is considered the most important program that runs on a computer because it manages all the hardware and software on it. It also controls the display/keyboard. </span>
Answer:
C. Process a risk acceptance for 2633 and remediate 3124.
Explanation:
There are various business risks. Some are inherited risks while other are risks are associated with nature of business. It is dependent on business owners that they want to accept risk or mitigate the risk. Risk acceptance is based on the strategy of the management. In the given scenario Accounting Prod Production has low risk 2633 which is accepted while 3124 is high risk which is remediated.
Answer:
I need to send black-and-white images in a print-ready format. Therefore, here are the steps I’d follow:
First, I’d shoot my images.
I’d use Photoshop to modify or correct the color of the images.
I’d convert the images into black-and-white versions.
I’d save the selected files in the TIFF format. This method will ensure that the images retain their quality and are print-ready.
I’d also convert these files into the JPEG format to get good-quality, low-resolution images for the client’s web publishing.
The Art Director requires images that they can modify, if required. Therefore, I’ll include files in the TIFF and PSD formats. These are open files and anyone can modify them.
I’d send all these saved copies to the magazine for their work.
Explanation:
just did it and it gave me this answer:)
Answer:
C++ code is given below
Explanation:
#include <iostream>
#include <cctype>
#include <string.h>
#include <cstring>
#include <sstream>
using namespace std;
struct Car {
public:
char reportingMark[5];
int carNumber;
string kind;
bool loaded;
string destination;
};
void input(Car *);
void output(Car *);
int main() {
Car *T = new Car;
input(T);
output(T);
delete T;
return 0;
}
void input(Car *T)
{
string str, s;
cout << " Enter the reporting mark as a 5 or less character uppercase string: ";
cin >> str;
for (int i = 0; i < str.length(); i++)
T->reportingMark[i] = toupper(str[i]);
cout << " Enter the car number: ";
cin >> T->carNumber;
cout << " Enter the kind: ";
cin >> T->kind;
cout << " Enter the loaded status as true or false: ";
cin >> s;
istringstream(s) >> boolalpha >> T->loaded;
if (T->loaded == true) {
cout << " Enter the destination: ";
cin.ignore();
getline(cin, T->destination);
}
else
T->destination = "NONE";
}
void output(Car *T)
{
cout << " Reporting Mark: " << T->reportingMark;
cout << " Car Number: " << T->carNumber;
cout << " Kind: " << T->kind;
cout << " Loaded Status: " << boolalpha << T->loaded;
cout << " Destination: " << T->destination << " ";
}
Answer:
(11n-5) / 12 is correct answer.
Explanation:
The Probability that key will match to nth term = 1/2
The Probability that key will match to n-1th term = 1/3
As all other probabilities are equal
The Total Probability that key matches to any of 1 to n-2 index = 1 - 1/2 - 1/3 = 1/6
The Probability that key matches to any of 1 to n-2 index = (1/6) / n-2 = (1/6)* (n-2))
Let P(i) = Probability that key matches to ith index.
The Average time complexity = 22 i=1 P(i) * i
The Average time complexity = 1/(6(n-2) * ( sum of 1 to n-2 ) + (n-1) / 3 + n/2
The Average time complexity = 1/(6(n-2) * ( n-2)*(n-1) / 2 + ( n-1) / 3 + n/2
The Average time complexity = 1/6 * (n-1)/2 + (n-1)/3 + n/2
The Average time complexity = (n-1)/12 + (n-1)/3 + n/2
The Average time complexity = (n-1 + 4 * n - 4 * 1 +6 * n)/12
The Average time complexity = 11n-5 / 12
so (11n-5) / 12 is correct answer.
