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bulgar [2K]
2 years ago
11

Which tab will you use to format data in cells?

Computers and Technology
2 answers:
kaheart [24]2 years ago
7 0

Answer:

It's home tab

Explanation:

vodka [1.7K]2 years ago
7 0

Answer:

B. Home tab

just got it correct :)

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How many searches should you complete when looking up information using online references?<br>​
antoniya [11.8K]
As many as it takes of course ;)
3 0
3 years ago
TCO 10) Voice packets could experience a significant _____ delay in routers and switches when sharing a network with data traffi
bonufazy [111]

This delay in routers is what we called Packetization delay or also called Accumulation delay. Packetization delay is the time required for the information to pass on the wires. The data rate of the links that passes thru the wires cause the delay.

8 0
3 years ago
⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val
timurjin [86]

Answer:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

8 0
2 years ago
A security technician is configuring a new firewall appliance for a production environment. The firewall must support secure web
Zielflug [23.3K]

Answer:

Explanation:

Based on the information provided in the question, the best rules that the technician should add to the firewall would be the following

Permit 10.10.10.0/24 0.0.0.0 -p tcp --dport 443

Permit 10.10.10.0/24 192.168.1.15/24 -p udp --dport 53

This is because port 443 is used for "Secure webs services" while UDP port 53 is used for queries and domain name resolution. Both of which are the main configurations that the security technician needs to obtain.

8 0
2 years ago
Show the output waveform of an AND gate with the inputs A, B,
NeX [460]

Answer:

Please see the attached image

Explanation:

6 0
3 years ago
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