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ipn [44]
3 years ago
5

The _________ specifies the station(s) for which the frame is intended. it may be a unique physical address, a group address, or

a broadcast address. select one:
a. da
b. sa
c. sfd
d. preamble
Computers and Technology
1 answer:
Hitman42 [59]3 years ago
7 0
<span>a. DA Given the question and available options, this looks like a question about Ethernet frames. So let's look at the options and see what makes sense. a. DA * DA - Destination address. And the question pretty much defines what it is. So this is the correct choice. b. SA * SA - Source address. This is the physical address that the frame was sent from. And since the question is mentioned who the frame is intended for, this is obviously the wrong choice. c. SFD * SFD - Start Frame Delimiter. This is a special CONSTANT sequence of bits that terminates the preamble. It's purpose is to indicate the start of the frame payload. Since it's a constant value, it does not indicate in any way where the frame is intended to go. So this is a bad choice. d. Preamble * Preamble - is a pattern of alternating 0 and 1 bits intended to synchronize receivers so that the rest of the frame is interpreted correctly. This too is a constant value and as such can't indicate the destination of the frame, so it's a bad choice.</span>
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Explanation:

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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.

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