Answer:
The answer is below
Explanation:
Suppose end system A wants to send a large file to end system B, the large file is first divided into smaller chunks. Each chunk is then assigned a header forming a packet. Multiple packets are being generated with each of the packet containing a unique destination address in its header.
When a packet arrives at the packet switch, the switch uses the unique destination address that is attached in the header of the packet to determine the link onto which the packet is to be forwarded.
Answer:
The correct code to the given question is
if ((counter % 10) == 0) // check the condition
{
System.out.println("Counter is divisible by ten: " + counter); // display
}
else // check the condition
{
System.out.println("Counter is not divisible by ten: " +counter); // display the //value
}
counter++; // increment the value of counter
Explanation:
Following are the description of code
- In the given question we have to check the condition that the given number is divisible by 10 or not .
- In the if block if the number is divisible by 10 then it print the value of number and increment the value of counter .
- In the else block if the number is not divisible by 10 then it print the value of number and increment the value of counter .
- It means the value of counter is increases in the if block as well as in the else block .So we have to remove the counter statement from there and place outside the if and else block .
Using our normal decimal numbering system, the base of a number system, for example 8, tells us two things
1. Each
digit is an integer that uses numbers from 0 to 7. There are 8 possible values
for a digit
2. We
multiply each digit by a power of 8 depending on the position of the digit.
If we use number 112 to (base 8), then;
<span>(1 x 8 to the power of 2)
+ (1 x 10 to the power of 1) + (2 x 8 to the power of 0). </span>
1. true
2. true
3. D
4. true
5. A
