Answer:
False
Explanation:
All Windows Server 2016 version are compatible with the common language runtime used in Microsoft .NET Framework and Microsoft Visual Studio .NET.
Also, Windows Server 2019 is compatible, along with Windows Server 2012 in all its versions
No regular Windows editions are compatible, but that is not strange, because they are not meant to be servers
Answer:
False
Explanation:
namespaces can be nested. That is we can have a hierarchy of namespaces.
For examples suppose we have a namespace top. Within this we have another namespace first. At the next level we have a namespace called second. Then we have a class MyClass as a member of this namespace second. Then the complete description of the class will be as follows:
top::first::second::MyClass
Answer:
The answer is 500 kbps
Explanation:
Consider the given data in the question.
R1 = 500 kbps
R2=2 Mbps
R3 = 1 Mbps
Now as it is mentioned that there is no other traffic in the network.
Thus,
throughput of the file = min {R1,R2,R3}
throughput of the file = min {500 kbps, 2 Mbps, 1 Mbps}
T/P of the file = 500 kbps
Answer:
The correct answer to the following question will be Option B (Control how well the e-learning artifacts interact technically with LMS).
Explanation:
SCORM seems to be a compilation of requirements and criteria for commodities used during e-learning. This describes the different ways to communicate here between information or data on the customer side as well as the run-time context.
- The specification also specifies how to bundle e-learning material for functionality, this is achieved by using the Bundle Exchange Format, an exchangeable ZIP file.
- It is essentially a manual that informs developers as well as content providers on how to make their apps or programs compliant with several other e-learning apps including applications.
The other three options are not related to the given scenario. So that option B is the right answer.
Answer:
The answer is below
Explanation:
a)
If the blocks of a file are not put together, that is for nonadjacent block of files to each other, the time (t) taken to read a 100 block file is:
t = (average seek + rotational latency + transfer time) * 100 block
Average seek = 5 msec, rotational latency = 10 msec and transfer time = 20 microsec = 0.02 msec
t = (5 + 10 + 0.02) * 100 = 1502 msec
b)
If the blocks of a file are put together, that is for adjacent block of files to each other, the time (t) taken to read a 100 block file is:
t = (seek time * mean interblock distance + rotational latency + transfer time) * 100 block
seek time = 100 microsec = 0.1 msec, rotational latency = 10 msec and transfer time = 20 microsec = 0.02 msec, mean interblock distance = 2 cylinders
t = (0.1*2 + 10 + 0.02) * 100 = 1022 msec
