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svlad2 [7]
2 years ago
5

Compiler is an example of​

Computers and Technology
1 answer:
maxonik [38]2 years ago
5 0

Answer:

System software

Explanation:

System software

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Einstein's famous equation states that the energy in an object at rest equals its mass times the squar of the speed of light. (T
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Answer:

The complete program is as follows:

m_str = input('Input m: ')

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Explanation:

This is an unchanged part of the program

m_str = input('Input m: ')

This converts m_str to float

mass = float(m_str)

This calculates the energy, e

e = mass * 300000000**2

This is an unchanged part of the program

print("e = ",e)

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⦁ 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.

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