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sertanlavr [38]
3 years ago
5

When looking at aggregated logs, you are seeing a large percentage of Windows hosts connecting to an Internet Protocol (IP) addr

ess outside the network in a foreign country. Why might this be worth investigating more closely
Computers and Technology
1 answer:
mario62 [17]3 years ago
7 0

Answer:

"It might demonstrate a malware infection" is the correct response.

Explanation:

  • Because once glancing at aggregate data log files, clients could perhaps actually listen to road conditions but instead correlation coefficient.  
  • For obvious reasons, whenever you see a significant portion of presenters designed to connect to something like a particular destination out beyond your channel, it might have been worthwhile to investigate quite tightly, although it might demonstrate malware.
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the answer is C. Insert

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State ONE (1) disadvantage of using computers to store and process<br> information.
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Answer:

it can be hacked

Explanation:

there is always a chance for computer data to be hacked

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The IT department has rejected the password you submitted for approval to join the network. Your suggestion was "bobbie." Why di
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On line two, you take a string, word, as input. Then, using that string...
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7

Explanation:

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8 0
3 years ago
g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
grin007 [14]

Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

3 0
3 years ago
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