Part 1 of 4 parts for this set of problems: Given an 4777 byte IP datagram (including IP header and IP data, no options) which is carrying a single TCP segment (which contains no options) and network with a Maximum Transfer Unit of 1333 bytes. How many IP datagrams result, how big are each of them, how much application data is in each one of them, what is the offset value in each. For the first of four datagrams: Segment

Fragment 1: size (1332), offset value (0), flag (1)

Fragment 2: size (1332), offset value (164), flag (1)

Fragment 3: size (1332), offset value (328), flag (1)

Fragment 4: size (781), offset value (492), flag (1)

Explanation:

The maximum = 1333 B

the datagram contains a header of 20 bytes and a payload of 8 bits( that is 1 byte)

The data allowed = 1333 - 20 - 1 = 1312 B

The segment also has a header of 20 bytes

the data size = 4777 -20 = 4757 B

Therefore, the total datagram = 4757 / 1312 = 4

some have built technology like spoofed call blockers into their networks, so there's nothing you would even have to install.

this happens because the number is known for scamming, they don't want it to me known as "scam likely." perhaps it's just a telemarketer. maybe it's an identity thief. it might even be a scam artist.  

either way, "scam likely" is probably not an old friend whose number you forgot that you saved. it's your wireless carrier's way of letting you know: answer at your own risk.

explanation:

to combat the seemingly rising epidemic of spam callers in the u.s., the nation's four major wireless carriers have been trying out new methods to reduce the number of unsolicited calls americans receive each day.

some have built technology like spoofed call blockers into their networks, so there's nothing you would even have to install.

hope this . mark me brainliest.

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