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maks197457 [2]
3 years ago
11

Which of the following information is okay to share on social networking site?

Computers and Technology
2 answers:
NNADVOKAT [17]3 years ago
8 0
C. your hobbies and interests
zvonat [6]3 years ago
4 0
C. Your hobbies and interests
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The transport layer protocol used by the tcp / ip suite that does not provide guarantees on ordering or confirmation of receipt
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What can be defined as in information technology environment?
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How do most database users perform data searches?
defon

Answer:

by using forms

Explanation:

Most of the database users perform the data searches by using the forms. A form can be used to enter, edit and display the data from the data source. Its a user interface in fact that fetches the data from the database. Reports have used the display the data for a certain type of user, and the viewing table does not look feasible to search from a very large database, and the databases are usually large. No calculation is required for searching the data, and we only need to write queries in the right syntax. Hence, here the correct option is by using forms.

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In an ipv4 address, what are the maximum number of bits that can be used to identify the network address
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5 0
3 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

  • Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW  = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

5 0
3 years ago
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