Answer:
D. All of the above are correct.
Explanation:
Asynchronous Transfer Mode is a fixed-length cell structure which allows cells to support voice, video and data. Voice data can be converted to packets and shared with large packet data. Because voice packets, encounter queuing delays, they ought to be of the same size or length.
ATM provides a route between two end points and that is why it can be said to be connection oriented. There is an ease of switching in hardware because of the fixed structure of the ATM. It also offers predictability regarding latency which could be either high or low. So, all of the above satisfy the mode of operation of the Asynchronous Transfer Mode.
It will help you present data with concise statements.
The INPUT code of the given programme in Java would be as given below:
In this first we are taking input of two integers j and k from the user and then performing the required operations over it
import java.util.
Scanner; public class LabProgram {
public static void main(String[ ] args) {
Scanner scnr = new Scanner(System.in);
int i,j;
i = scnr.nextInt();
j = scnr.nextInt();
if(i<=j) {
int k = i;
while (k <= j) {
System.out.print(k + " ");
k = k + 5;
}
System.out.println();
}
else{
System.out.println("Second integer can't be less than first.");
}
}
}
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A program that loops and performs powers of 5 on each loop.In each lap the operator is asked IN CODE C++ the C++ Program for the given hassle announcement is as follows:#include the use of namespace std.
<h3>What is the energy of a variety of?</h3>
The energy of a variety of says how usually to apply the range in a multiplication. Powers also are known as Exponents or Indices. For example, eight^2 might be known as “eight to the energy 2” or “eight to the second one energy”, or simply “eight square.
- C++ Program for the given hassle announcement is as follows:
- #include
- the use of namespace std;
- int main() > base;
- cout << "nEnter the exponent: "; //asks the user for keyboard input for exponent
- cin>>exponent;
- for (int i = 1; i <=exponent; i++) //iterate {through|thru|via">via 1 until exponent
- {
- num=num*base;
- //base is multiplied (e.g. 2*2*2.....n) until n, wherein n is exponent
- }
- cout <<"n"<strength of "<go back 0;
- }
- Sample Output:
- Stdin Inputs
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two people have to answer the question and at the bottom of the question you'll see a crown and you just click on that to give them brainliest
