COMPUTERS AND TECHNOLOGY COLLEGE

Which orientation is wider than it is tall?Portrait
Macro
Shutter
Landscape

Answers

Answer 1
Answer:

Answer:

Landscape

Explanation:

Landscape orientation is wider than it is tall.
Answer 2
Answer:

Answer:

Landscape

Explanation:

Related Questions

In this project you will write a set of instructions (algorithm). The two grids below have colored boxes in different locations. You will create instructions to move the colored boxes in grid one to their final location in grid two. Use the example to help you. The algorithm that you will write should be in everyday language (no pseudocode or programming language). Write your instructions at the bottom of the page.Example: 1. Move the orange box 2 spaces to the right.2. Move the green box one space down. 3. Move the green box two spaces to the left.Write your instructions. Review the rubric to check your final work.Rules: All 6 colors (red, green, yellow, pink, blue, purple) must be move to their new location on the grid. Block spaces arebarriers. You cannot move through them or on them – you must move around them
_ is an important management tool that is used to monitor and manage that contract and/or project
Which statement prints "hi" on the screen?(a) puts("hi");(b) put "hi";(c) puts "hi";(d) none of the above
Express 278910 in binary – Use the technique of subtracting powers of 2
How many times will the following loop display "Looping again!"? for(int i=0; i<=20; i++) cout << "Looping again!" << endl;

9 Which of these words is used to begin a conditional statement? ​

Answers

Answer:

The word used to begin a conditional statement is if.

Explanation:

In the syntaxes of most (if not all) programming languages, the word if is used to begin any conditional statement.

Take for instance:

Python:

if a == b:

   print("Equal")

C++:

if(a == b)

cout<<"Equal";

Java:

if(a==b)

System.out.print("Equal")

The above code segments in Python, C++ and Java represents how conditional statements are used, and they all have a similarity, which is the "if" word

What types of messages flow across an SDN controller’s northbound and southbound APIs? Who is the recipient of these messages sent from the controller across the southbound interface, and who sends messages to the controller across the northbound interface?

Answers

The types of messages which flow across an SDN controller northbound and southbound APIs are:

  • Messages that helps to develop flow tables
  • Messages that helps to interact between the controller and network control applications
  • Messages which help for the up-to-date view of the network's state, etc.

The recipient of these messages sent from the controller across the southbound interface, and who sends messages to the controller across the northbound interface are:

  • Controlled devices are the recipient
  • Network control applications send the messages

What is Communication?

This refers to the exchange of information between different people or computers using a medium which also gives feedback.

Read more about communication here:
brainly.com/question/25793182

Answer and Explanation:

Messages flow across an SDN controller's:  

Northbound APIs:  

• Messages which help in read/write state of the network and developing flow tables within the  

state management layer.  

• Notifications for the state-change events.  

• The interaction between the controller and network control applications is done through the  

northbound interface.  

• Network control applications send messages to the controller.  

Southbound APIs:  

• Messages which help for the up-to-date view of the network's state like message for the  

attached link has gone up or down, new devices are joined the network, or indications of the  

device is up or down.  

• Controller's southbound interface is the communication among the controller and the controlled  

devices. Controlled devices are the recipients of the messages sent form the controller.  

From an IT perspective, which of the following best describes BI and BI apps?a. Stand-alone
b. Support a specific objective
c. A collection of ISs and technologies
d. Web-based systems designed for for-profits

Answers

Answer:a)Stand-alone

Explanation: Stand-alone application is the application that is found on the system of every client.In accordance with the IT section, Business intelligence can be transferred into stand-alone application .This helps in the development of the essence of the system at an independent level.

Other options are incorrect because supporting a certain factor will not make it independent, cannot act as the group of ISs technology or web system for gaining profit.Thus,the correct option is option(a).

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.

Express the worst case run time of these pseudo-code functions as summations. You do not need to simplify the summations. a) function(A[1...n] a linked-list of n integers) for int i from 1 to n find and remove the minimum integer in A endfor endfunction

Answers

Answer:

The answer is "O(n2)"

Explanation:

The worst case is the method that requires so many steps if possible with compiled code sized n. It means the case is also the feature, that achieves an average amount of steps in n component entry information.

  • In the given code, The total of n integers lists is O(n), which is used in finding complexity.
  • Therefore, O(n)+O(n-1)+ .... +O(1)=O(n2) will also be a general complexity throughout the search and deletion of n minimum elements from the list.

(5 pt.) The name of a variable in the C programming language is a string that can contain uppercase letters, lowercase letters, digits, or underscores. Furthermore, the first character in the string must be a letter, either uppercase or lowercase, or an underscore. If the name of a variable is determined by its first 8 characters, how many different variables can be named in C

Answers

Answer:

The different variable in C is 21213316700.

Explanation:

Given value:

Total value  = letters + underscore value

Total value  = 52 + 1

Total value =53

choice for first character = 53 letters +10 digits

first character = 63

choice for remaining characters

So,

Variable number With one 1 character = 53

Variable number With 2 character  = 53 × 63

Variable number With 3 character = 53 × 63²

Variable number With 4 character  = 53 × 63³

.

.

.

Variable number With 7 character  = 53 * 63^(7)

Total difference variable = 53 + 53 × 63+ 53 × 63²+ 53 × 63³+....+ 63^(7)

Total difference variable = 53(1 + 63 + 63²+ 63³+ .... + 63^(7))

Formula:

1+x+x^2+x^3+x^4+........x^n\n\n\Rightarrow (x^((n+1)) -1)/(x-1)

Total difference variable

53 * (63^(7+1) -1 )/(63-1)\n\n53 * (63^(8) -1 )/(63-1) \n\n \therefore 63^8 = 2.48 * 10^(14)\n\n53 * (2.48 * 10^(14) -1 )/(62)\n\n53 * (1.48 * 10^(14))/(62)\n\n21213316700
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