Miriam Is a network administrator. A few employees want to access sensitive Information stored on a backup device. She wants to
give access
rights these employees. Which option will she use?
ОА.
remote desktop
ОВ.
monitoring service
Oc.
user accounts
OD.
print service
OE.
directory service
rights these employees. Which option will she use?
ОА.
remote desktop
ОВ.
monitoring service
Oc.
user accounts
OD.
print service
OE.
directory service
1 answer:
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Answer:
Here is the C++ program:
#include <iostream> //to use input output functions
#include <fstream> //to manipulate files
using namespace std; //to identify objects like cin cout
int main(){ //start of main function
ifstream file; //creates an object of ifstream
file.open("random.txt"); //open method to open random.txt file using object file of ifstream
int numCount = 0; //to store the number of all numbers in the file
double sum = 0; //to store the sum of all numbers in the file
double average = 0.0; //to store the average of all numbers in the file
int number ; //stores numbers in a file
if(!file){ //if file could not be opened
cout<<"Error opening file!\n"; } //displays this error message
while(file>>number){ //reads each number from the file till the end of the file and stores into number variable
numCount++; //adds 1 to the count of numCount each time a number is read from the file
sum += number; } //adds all the numbers and stores the result in sum variable
average = sum/numCount; //divides the computed sum of all numbers by the number of numbers in the file
cout<<"The number of numbers in the file: "<<numCount<<endl; //displays the number of numbers
cout<<"The sum of all the numbers in the file: "<<sum<<endl; //displays the sum of all numbers
cout<<"The average of all the numbers in the file: "<< average<<endl; //displays the average of all numbers
file.close(); } //closes the file
Explanation:
Since the random.txt is not given to check the working of the above program, random.txt is created and some numbers are added to it:
35
48
21
56
74
93
88
109
150
16
while(file>>number) statement reads each number and stores it into number variable.
At first iteration:
35 is read and stored to number
numCount++; becomes
numCount = numCount + 1
numCount = 1
sum += number; this becomes:
sum = sum + number
sum = 0 + 35
sum = 35
At second iteration:
48 is read and stored to number
numCount++; becomes
numCount = 1+ 1
numCount = 2
sum += number; this becomes:
sum = sum + number
sum = 35 + 48
sum = 83
So at each iteration a number is read from file, the numCount increments to 1 at each iteration and the number is added to the sum.
At last iteration:
16 is read and stored to number
numCount++; becomes
numCount = 9 + 1
numCount = 10
sum += number; this becomes:
sum = sum + number
sum = 674 + 16
sum = 690
Now the loop breaks and the program moves to the statement:
average = sum/numCount; this becomes:
average = 690/10;
average = 69
So the entire output of the program is:
The number of numbers in the file: 10 The sum of all the numbers in the file: 690 The average of all the numbers in the file: 69
The screenshot of the program and its output is attached.
