Answer:
The time elapsed is 0.017224 s
Solution:
As per the question:
Analog signal to digital bit stream conversion by Host A =64 kbps
Byte packets obtained by Host A = 56 bytes
Rate of transmission = 2 Mbps
Propagation delay = 10 ms = 0.01 s
Now,
Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.
Time taken to generate and convert all the bits into digital signal is given by;
t = 
t = 
(Since, 1 byte = 8 bits)
t = 7 ms = 0.007 s
Time Required for transmission of the packet, t':
Now, the time elapse between the bit creation and its decoding is given by:
t + t' + propagation delay= 0.007 + 
+ 0.01= 0.017224 s
Task view is the option that you would not see on a Windows 10 Start menu, whereas if you click the Windows button, you can see tiles with different applications, all of your apps in the form of tiles you can distribute according to your own taste, and power button where you can choose whether you want to shut down your computer, put it to sleep, or log out.
Answer:
translation is used for interpretation
Answer:
- public class FindDuplicate{
-
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
-
- int n = 5;
- int arr[] = new int[n];
-
- for(int i=0; i < arr.length; i++){
- int inputNum = input.nextInt();
- if(inputNum >=1 && inputNum <=n) {
- arr[i] = inputNum;
- }
- }
-
- for(int j =0; j < arr.length; j++){
- for(int k = 0; k < arr.length; k++){
- if(j == k){
- continue;
- }else{
- if(arr[j] == arr[k]){
- System.out.println("True");
- return;
- }
- }
- }
- }
- System.out.println("False");
- }
- }
Explanation:
Firstly, create a Scanner object to get user input (Line 4).
Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).
Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.
If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".
