ATX
microATX (smaller version of ATX)
mini-ITX (smaller version of microATX)
Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.
True
<u>Explanation:</u>
In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.
Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.
Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.
Answer:
The answer to this question can be defined as follows:
Explanation:
Its Permute-with-all method, which doesn't result in a consistent randomized permutation. It takes into account this same permutation, which occurs while n=3. There's many 3 of each other, when the random calls, with each one of three different values returned and so, the value is= 27. Allow-with-all trying to call possible outcomes as of 3! = 6
Permutations, when a random initial permutation has been made, there will now be any possible combination 1/6 times, that is an integer number m times, where each permutation will have to occur m/27= 1/6. this condition is not fulfilled by the Integer m.
Yes, if you've got the permutation of < 1,2,3 > as well as how to find out design, in which often get the following with permute-with-all chances, which can be defined as follows:
Although these ADD to 1 none are equal to 1/6.
Answer:
"Every time you try to run your new game, it crashes after displaying the opening screen"
"After a recent Windows update, you can no longer access the internet on your laptop"
Explanation:
