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mrs_skeptik [129]
4 years ago
8

A technician with a PC is using multiple applications while connected to the Internet. How is the PC able to keep track of the d

ata flow between multiple application sessions and have each application receive the correct packet flows
Computers and Technology
2 answers:
Andre45 [30]4 years ago
7 0

Answer:

The data flow is being tracked based on the source port number utilized by each application

Explanation:

Every application has a source port number which is usually generated randomly and used for the purpose of keeping track of every session of connection to the internet on an individual basis. In this way, each application is required to use a source code unique to an application to provide concurrent communication from a number of applications via the internet.

kvv77 [185]4 years ago
5 0

Answer:  With the tracking via source ports.

Explanation:

Hi, each application has a unique source port number randomly generated, it allows the pc to track the data flow.

Because of this a pc can send multiple HTTP service requests to a web server at the same time, and these requests can be tracked based on the source ports.

Feel free to ask for more if needed or if you did not understand something.

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Answer:

RAM Slots

Explanation:

RAM is most probably in the wrong slots.

Check BIOS

Check with Motherboard User Manual.

7 0
3 years ago
Out of a list of the values 8, 3, 30, 5, 15, and 6, what result would the MIN function return?
kondor19780726 [428]
Probably the minimum number in the given set so 5.
6 0
3 years ago
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How can you enter Task Manager in Windows? Select 3 options. press Ctrl + Shift + Tab press Ctrl+Alt+Delete and then click Task
Sophie [7]

Answer:

Please find the 3 options below.

Explanation:

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5 0
3 years ago
A 10 Superscript negative 910−9​-F capacitor ​(11 nanofaradnanofarad​) is charged to 5050 V and then disconnected. One can model
vredina [299]

Answer:

Explanation:

i) On a cold day, Capacity = (66 X 910 X 10∧-9) F

Charge, Q = (66 X 910 X 10∧-9) X 5050 = 0.30 Columb

Current, I= 5050/131013 = 0.039 Amp (where Resistance = 131013 ohms)

Time, t= 0.30/0.039 = 7.7 seconds

ii) On a humid day, Capacity = (77 X 910 X 10∧-9) F

Charge, Q = ( 77 X 910 X 10∧-9) X 5050 = 0.35 Columb

Current, I= 5050/6106 = 0.83 Amp ( where Resistance = 6106 ohms)

Time, t= 0.35/0.83 = 0.42 h or  0.42 X 60 mins = 25 mins

7 0
3 years ago
DO NOT ANSWER FOR JUST POINTS
Elza [17]

Answer:

E.

a medium dashed line

Explanation:

I always use medium hehe so I think that'll work.

5 0
3 years ago
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