Describe how the Internet Protocol (IP) allows devices to easily connect and communicate on the Internet.

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Answer 1
Answer:

Internet Protocol (IP) are guideline about how data should be sent across the internet.

Internet Protocol (IP) work hand in hand with packet as they are connected to every packet which therefore means that any internet users who connect to the internet will be allocated with a unique Internet Protocol (IP)  address.

This  Internet Protocol (IP)  address will  enables the user devices to connect and communicate across the internet with ease once the packets are connected to the each Internet Protocol (IP)  address allocated to them.

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one of the 4 vs of big data that refers to uncertainty due to data inconsistency and incompleteness, ambiguities, latency, deception, and model approximations is . select one: veracity volume validity variety

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one of the 4 vs of big data that refers to uncertainty due to data inconsistency and incompleteness, ambiguities, latency, deception, and model approximations is veracity.

What is veracity?

Conformity with truth or fact : accuracy. : devotion to the truth : truthfulness. 3. /vəˈraes.ə.ti/ the quality of being true, honest, or accurate: Doubts were cast on the veracity of her alibi. Synonyms.  Veracity is the quality of being true or the habit of telling the truth. He was shocked to find his veracity questioned. Veracity refers to the quality of the data that is being analyzed. High veracity data has many records that are valuable to analyze and that contribute in a meaningful way to the overall results. Low veracity data, on the other hand, contains a high percentage of meaningless data. This type of source is the starting point for any historian-researcher. Some written sources are forgeries.

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To define constructors and member functions outside of a class's original scope, the operator can be used.

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Answer: Scope resolution operator(::)

Explanation: A member function and the constructor can be called within the function easily but for the execution of the these components outside the class , a special operator is required to call the functions. The scope resolution operator(::) preceding with the name of class is thus used for defining of the function outside class.This operator maintains the cope of the function and constructor outside the class.

Design an algorithm to find all the common elements in two sorted lists of numbers. For example, for the lists (2, 5, 5, 5)
and (2, 2, 3, 5, 5, 7), the output should be (2, 5, 5).

Answers

Answer:

while list1 not empty AND list2 not empty {

    if elements at leftmost positions are equal {

        pop element from list1

        pop element from list2

        add element to result_list

    } else {

        pop lowest value from list1 or list2

    }

}

Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit, and worst-fit algorithms place processes of size 200 MB, 15 MB, 185 MB, 75 MB, 175 MB, and 80 MB (in order)? Indicate which—if any—requests cannot be satisfied. Comment on how efficiently each of the algorithms manages memory.

Answers

Using First fit algorithm:

  • P1 will be allocated to F4. With this, F4 will have a remaining space of 5MB from (205 - 200).
  • P2 will be allocated to F1. With this, F1 will have a remaining space of 85MB from (100 - 15).
  • P3 will be allocated F5. With this, F5 will have a remaining space of 115MB from (300 - 185).
  • P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
  • P5 will be allocated to F6. With this, F6 will have a remaining space of 10MB from (185 - 175).
  • P6 will be allocated to F2. With this, F2 will have a remaining space of 90MB from (170 - 80).

Using Best-fit algorithm:

  • P1 will be allocated to F4. With this, F4 will have a remaining space of 5MB from (205 - 200).
  • P2 will be allocated to F3. With this, F3 will have a remaining space of 25MB from (40 - 15).
  • P3 will be allocated to F6. With this, F6 will have no remaining space as it is entirely occupied by P3.
  • P4 will be allocated to F1. With this, F1 will have a remaining space of of 25MB from (100 - 75).
  • P5 will be allocated to F5. With this, F5 will have a remaining space of 125MB from (300 - 175).
  • P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).
  • 100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

Using Worst-fit algorithm:

  • P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
  • P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
  • P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
  • P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
  • P5 will not be allocated to any of the available space because none can contain it.
  • P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

Properly labeling the six different processes would be:

  • 200MB (P1),
  • 15MB (P2),
  • 185MB (P3),
  • 75MB (P4),
  • 175MB (P5)
  • 80MB (P6).

Best fit algorithm is the best as the name suggests, while the worst fit algorithm is the worst as not all memory is allocated

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Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
  3. P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
  4. P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
  5. P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
  3. P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
  4. P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
  5. P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
  6. P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

  1. P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
  2. P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
  3. P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
  4. P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
  5. P5 will not be allocated to any of the available space because none can contain it.
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

Which of the following is NOT a typical action of the catch block?Select one:
a. Throwing the exception
b. Completely handling the exception
c. Rethrowing the same exception for the calling environment
d. Partially processing of the exception

Answers

Answer:

A.

Explanation:

because its throwing the exception

What is Interface in computers

Answers

Answer:

An interface could be the design of the homescreen or the way a shell of a computer is designed. I could be wrong tho......

Explanation: