Waiting time for a 100 Mbps broadcast channel is 0.512 ms
Recall that with the CSMA/CD protocol, the adapter waits K * 512 bit times after a collision,
where K is drawn randomly. For K = 100,
The one bit time for 10 Mbps is 1/10000000 s = 0.0001 ms
The one bit time for 100 Mbps is 1/100000000 s = 0.00001 ms
Waiting time for a 10 Mbps broadcast channel is K*512*0.0001= 5.12 ms
Waiting time for a 100 Mbps broadcast channel is K*512*0.00001= 0.512 ms
<h3>
What is a collision in Ethernet?</h3>
- A collision happens on a half-duplex Ethernet network when two devices on the same network attempt to communicate data at the exact same time.
- The two transmitted packets are "collapsed" by the network, which results in the network discarding both of them.
- On Ethernets, collisions are unavoidable.
- A collision is a momentary interaction between two bodies or more than two bodies at once that modifies the motion of the bodies involved as a result of the internal forces at work.
- Collisions entail the application of force (there is a change in velocity).
Learn more about collision in Ethernet: brainly.com/question/14123270
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Answer:
- #include <iostream>
- using namespace std;
- int main()
- {
- int highest = 0;
- int score;
- do{
- cout<<"Input a score: ";
- cin>>score;
-
- if(score > highest){
- highest = score;
- }
- }while(score >= 0);
-
- cout<<highest;
- return 0;
- }
Explanation:
Firstly, create a variable highest and initialize it with zero (Line 5). Next, create a do while loop (Line 7 - 14). Within the loop prompt user to input a score (Line 8-9) and if the current score is higher than the highest variable, assign the score to highest variable (Line 11 - 13).
After finishing the loop when user put in any negative value, the program shall be able to print out the highest input score (Line 16).
Answer:
i mean, theres not much strategy to the game at all, its just being able to shoot from hard spots, and obviously have good consistentsy
Explanation:
Answer:
The answer is 2988 bits per seconds
Explanation:
Solution
Given that:
Bandwidth (B) = 300Hz
Signal to noise ratio (SNR) = 30dB
Now let C = channel capacity
Using Deubel formula we have the following:
SNR db = 10* log (SNR)
30 =10* log (SNR)
So,
SNR =log^⁻1 (3)
SNR = (10)^3
SNR = 1000
Thus SNR =1000
Now
Applying the Shannon's equation
C = B * log₂ ( 1 + SNR)
C = 300 * log₂ ( 1 + 1000)
C = 300 * log₂ (1001)
C = 300 * 9.96
C =2988
Hence the capacity of the channel for teleprinter channel is 2988 bits per second.
Answer: a)Purchased from a vendor and gives you the right to use the software but is not yours to own.
Explanation: Proprietary software is the software that can be used by the user but it has copyright of the vendor/holder in any case.There are some certain condition under which the user can use this software.The distribution of the software is not permitted in this case. No modification or manipulation can be done by the user.
Other options are incorrect because it is not free as its purchased from vendor ,not freely available rather guided by owner and can be used over certain period of time.Thus , the correct option is option(a).
