What are the similarities and differences between the editor-in-chief, managing editor, assignment editor, and copyeditor? Your response should be at least 150 words.

Answers

Answer 1
Answer:

Answer:

Editor. An editor is the individual in charge of a single publication. It is their responsibility to make sure that the publication performs to the best of its ability and in the context of competition. A managing editor performs a similar role, but with greater responsibility for the business of the publication.

Explanation:


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Problem 1 (10%): You are asked to design a database for an auto-shop that satisfies the following requirements: a. Each customer has a unique customer identification number, a name, address, day time telephone number, evening telephone number. The customer name includes first name, middle initial and last name. The address includes street address, city, state, and zip code. The telephone numbers include area code and the number. b. Each car type has a unique type name, the make, model, and year of make. c. The car has a unique license number, a car type, color, type of transmission, and the customer who brings the car to the auto-shop. d. The database keeps records of repairs. The repair record includes the date of the repair, the car that the repair was done, the name of the engineer who did the repair, and the cost of the repair.
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Preserving confidentiality, integrity, and availability is a restatement of the concern over interruption, interception, modification, and fabrication. i. Briefly explain the 4 attacks: interruption, interception, modification, and fabrication.
ii. How do the first three security concepts relate to these four attacks

Answers

Answer and Explanation:

I and II answered

Interruption: interruption occurs when network is tampered with or communication between systems in a network is obstructed for illegitimate purposes.

Interception: interception occurs when data sent between systems is intercepted such that the message sent to another system is seen by an unauthorized user before it reaches destination. Interception violates confidentiality of a message.

Modification: modification occurs when data sent from a system to another system on the network is altered by an authorized user before it reaches it's destination. Modification attacks violate integrity, confidentiality and authenticity of a message.

Fabrication: fabrication occurs when an unauthorized user poses as a valid user and sends a fake message to a system in the network. Fabrication violates confidentiality, integrity and authenticity.

Is the willingness to put a customer’s needs above ones own needs and to go beyond a job description to achieve customer satisfaction

Answers

Answer:

customer-serious orientation

Explanation:

Customer-serious orientation can be defined as the willingness to put a customer’s needs above ones own needs and to go beyond a job description to achieve customer satisfaction.

This ultimately implies that, a customer-serious oriented business firm or company puts the needs, wants and requirements of its customers first without considering their own needs in a bid to satisfy and retain them.

Hence, customer-serious orientation requires the employees working in an organization to show and demonstrate positive attitudes and behaviors at all times.

g A peer-to-peer PLC network: has no master PLC. uses the token passing access control scheme. each device is identified by an address. all of these.

Answers

Answer:

All of these

Explanation:

A peer-to-peer network is a type of computer network that computer devices directly to each other, sharing information and other resources on the network.

It does not have a centralized database, there are no master or slave as each device in the network is a master on its own, and the devices all have unique addresses assigned statically for identification.

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

Read more about memory partitions here:
brainly.com/question/12726841

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.

Write an if statement that assigns 100 to x when y is equal to 0.

Answers

Answer:

if(y==0)

{

   x=100;

}

Explanation:

The above written if statement is for assigning 100 to x when the value of y is equal to 0.To check the value of the y I have used equal operator == which returns true when the value on it's left side is equal to the value to it's right else it returns false and for assigning the value to y I have used assignment operator =.

First it will be checked that the value y is equal to 0.If the then it return true means the if statement will execute.Inside if statement 100 is assigned to x.

Which one of these are a valid IPv4 address? Check all that apply.A. 1.1.1.1
B. 345.0.24.6
C. 54.45.43.54
D. 255.255.255.0

Answers

The valid IPv4 addresses are A. 1.1.1.1, C. 54.45.43.54, and  D. 255.255.255.0

Valid IPv4 addresses are 32 bits in size and normally contain three periods and four octets like 1.1.1.1, and the value can be any number from zero to 255.

IPv4 means Internet Protocol version 4.  It is a set of address that identifies a network interface on a computer.

Thus, the only invalid address according to IPv4 standards is B. 345.0.24.6, with others regarded as valid IPv4 addresses.

Learn more about IPv4 here: brainly.com/question/19512399