The data components of a class that belong to every instantiated object are the class's ___

Match the vocabulary word to the accurate definition. A software program that enables you to search for, interact with, and retr

SashulF [63]

Answer:

A software program that enables you to search for, interact with, and retrieve information resources: Search Engine

A method by which data is transferred from one computer to another through the Internet Data Transfer

Coordinates actions between the user interface and the rendering engine: Operating Systems

Handles requests for information resources from networked servers: SERVER

Includes web pages, images, videos, music files, and other forms of content WEBSITE

Enables a browser to save browsing information Advanced Settings

Includes an address bar, Back and Forward buttons, bookmarking menu, and other useful features Web browser

Displays content on the screen. Monitor

Explanation:

If you enable the browser to remember the browser experience, then you certainly need to go inside the advanced settings and from there you can check remember last 7 days and other options. Content is displayed on Monitor, And it is the website that shows images, video, text etc. The information resources are being controlled through the server, and Data transfer is the transfer of the data from one computer to other via a proper network. And the Data is collected, interacted as well as retrieve information via the search engines. And operating system is definitely the interface, and an educated one in between user interface and the rendering engine.

4 0

2 years ago

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A school has 100 lockers and 100 students. All lockers are closed on the first day of school. As the students enter, the first s

kati45 [8]

Solution :

public class NewMain {

public_static void main_(String[] args) {

boolean[] _locker = new boolean_[101];

// to set all the locks to a false NOTE: first locker is the number 0. Last locker is the 99.

for (int_i=1;i<locker_length; i++)

{

locker[i] = false;

}

// first student opens all lockers.

for (int i=1;i<locker.length; i++) {

locker[i] = true;

}

for(int S=2; S<locker.length; S++)

{

for(int k=S; k<locker.length; k=k+S)

{

if(locker[k]==false) locker[k] = true;

else locker[k] = false;

}

for(int S=1; S<locker.length; S++)

{

if (locker[S] == true) {

System.out.println("Locker " + S + " Open");

}

/* else {

System.out.println("Locker " + S + " Close");

} */

}

5 0

2 years ago

a. a large central network that connects other networks in a distance spanning exactly 5 miles. b. a group of personal computers

ki77a [65]

Complete Question:

A local area network is:

Answer:

b. a group of personal computers or terminals located in the same general area and connected by a common cable (communication circuit) so they can exchange information such as a set of rooms, a single building, or a set of well-connected buildings.

Explanation:

A local area network (LAN) refers to a group of personal computers (PCs) or terminals that are located within the same general area and connected by a common network cable (communication circuit), so that they can exchange information from one node of the network to another. A local area network (LAN) is typically used in small or limited areas such as a set of rooms, a single building, school, hospital, or a set of well-connected buildings.

Generally, some of the network devices or equipments used in a local area network (LAN) are an access point, personal computers, a switch, a router, printer, etc.

4 0

2 years ago

8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101

Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

= 64 + 32 + 16 + 8 + 4 + 2 + 1

= 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

= 128 + 64 + 32 + 16

= 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

= 64 + 32 + 4 + 2 + 1

= 103

01111101 = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

= 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

= 64 + 32 + 16 + 8 + 4 + 1

= 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 10101111 11000000 11111000 00011101 to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 223

10110000 = 1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

= 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 247

11000111 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

= 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

= 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0

2 years ago

Sarah works in a coffee house where she is responsible for keying in customer orders. A customer orders snacks and coffee, but l

Maru [420]

Answer:

Stack

Explanation:

Stack is a linear data structure that follows a particular order in the way an operation is done or sequence a job is completed.

It uses either LIFO ( Last In, First Out) which is also known as first come first served sequence or FILO (First In, Last Out) sequence.

There are several real life examples of which we can use the example of replacing the snack items Sarah brought for the customer.

If Sarah used the LIFO method, it means she replaced the snack items first ontop of the already existing snack items that's why there is a mismatch.

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2 years ago

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The data components of a class that belong to every instantiated object are the class's ____ variables.