Answer: b) R1 will connect to a physical Ethernet link, with the other end of the cable connected to a device at the WAN service provider point of presence.
.d) R1 will forward data link frames to R2 using an Ethernet header/trailer.
Explanation:
Here we have to understand what is MLPS. MLPS is a protocol which identifies the shortest route for the transfer of messages between routers instead of the longest route.
Here we are given a WAN which is a wide area network. Using layer 2 Ethernet service the frames are transmitted across the routers within the WAN.
Option A is incorrect as as connecting the other end of the cable to the R2 would reduce the functionality of the WAN. Option C is incorrect as HDLC header trailer has no effect.
Option B and D are correct as Ethernet header/ trailer has the same size every frame and to maintain the functionality of the WAN one end must be connected to point of presence(PoP).
Answer:
import java.util.Scanner;
public class num8 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter month's budget");
double monthBudget = in.nextDouble();
double totalExpenses = 0.0;
double n;
do{
System.out.println("Enter expenses Enter zero to stop");
n = in.nextDouble();
totalExpenses += n;
}while(n>0);
System.out.println("Total expenses is "+totalExpenses);
System.out.println("The amount over your budget is "+ Math.abs(monthBudget-totalExpenses));
}
}
Explanation:
- Using Java programming language
- Prompt user for month's budget
- Use Scanner class to receive and store the amount entered in a variable
- Use a do while loop to continuously request user to enter amount of expenses
- Use a variable totalExpenses to add up all the expenses inside the do while loop
- Terminate the loop when user enters 0 as amount.
- Subtract totalExpenses from monthBudget and display the difference as the amount over the budget
Answer:
The recursion function is as follows:
def raise_to_power(num, power):
if power == 0:
return 1
elif power == 1:
return num
else:
return (num*raise_to_power(num, power-1))
Explanation:
This defines the function
def raise_to_power(num, power):
If power is 0, this returns 1
if power == 0:
return 1
If power is 1, this returns num
elif power == 1:
return num
If otherwise, it calculates the power recursively
else:
return (num*raise_to_power(num, power-1))
Answer:
yes i agree with that the answer is in fact true
Explanation:
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Answer:
A
Issues can lead reoccuring costomers to not want to come back, resulting in this issue.