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11Alexandr11 [23.1K]
2 years ago
11

Drag each label to the correct location on the image.

Computers and Technology
1 answer:
pantera1 [17]2 years ago
3 0

Answer:

1.computer-based  training modules  is an example of hierarchical technique.

Explanation:

Computer based training modules are the example of hierarchical model or technique of website. In this technique courses or modules are linked with each other in hierarchical structure. To start some advance module we should learn some basic modules first.

2.school website is an example of Linear Structure of Website.

Explanation:

In linear structure of website, pages are linked in a sequence. We can follow the links to find some information in sequential manner.

3.bookstore  website is an example of Linear Structure of Website.

Explanation:

In linear structure of website, pages are linked in a sequence. We can follow the links to find some information in sequential manner.

4.single-product  websites is an example of Linear Structure of Website.

Explanation:

In linear structure of website, pages are linked in a sequence. We can follow the links to find some information in sequential manner.

5.county library  website is an example of Webbed technique.

Explanation:

Computer based training modules are the example of webbed model or technique of website. In this technique many pages are connected or linked to many other pages and each page can access any other page. This is a complex structure of website.

6.online gift store  website is an example of Linear Structure of Website.

Explanation:

In linear structure of website, pages are linked in a sequence. We can follow the links to find some information in sequential manner.

7.federal government  website is an example of Webbed technique.

Explanation:

Computer based training modules are the example of webbed model or technique of website. In this technique many pages are connected or linked to many other pages and each page can access any other page. This is a complex structure of website.

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Write a method called justFirstAndLast that takes a single String name and returns a new String that is only the first and last
NemiM [27]

Answer:

Answer is in the attached screenshot.

Explanation:

Using regex to split a given input string via whitespace, then returns the first and last element of the array.

6 0
2 years ago
Modify the TimeSpan class from Chapter 8 to include a compareTo method that compares time spans by their length. A time span tha
My name is Ann [436]

Answer:

Check the explanation

Explanation:

Here is the modified code for

TimeSpan.java and TimeSpanClient.java.

// TimeSpan.java

//implemented Comparable interface, which has the compareTo method

//and can be used to sort a list or array of TimeSpan objects without

//using a Comparator

public class TimeSpan implements Comparable<TimeSpan> {

   private int totalMinutes;

   // Constructs a time span with the given interval.

   // pre: hours >= 0 && minutes >= 0

   public TimeSpan(int hours, int minutes) {

        totalMinutes = 0;

        add(hours, minutes);

   }

   // Adds the given interval to this time span.

   // pre: hours >= 0 && minutes >= 0

   public void add(int hours, int minutes) {

        totalMinutes += 60 * hours + minutes;

   }

   // Returns a String for this time span such as "6h15m".

   public String toString() {

        return (totalMinutes / 60) + "h" + (totalMinutes % 60) + "m";

   }

   // method to compare this time span with other

   // returns a negative value if this time span is shorter than other

   // returns 0 if both have same duration

   // returns a positive value if this time span is longer than other

   public int compareTo(TimeSpan other) {

        if (this.totalMinutes < other.totalMinutes) {

            return -1; // this < other

        } else if (this.totalMinutes > other.totalMinutes) {

            return 1; // this > other

        } else {

            return 0; // this = other

        }

   }

}

// TimeSpanClient.java

public class TimeSpanClient {

   public static void main(String[] args) {

        int h1 = 13, m1 = 30;

        TimeSpan t1 = new TimeSpan(h1, m1);

        System.out.println("New object t1: " + t1);

        h1 = 3;

        m1 = 40;

        System.out.println("Adding " + h1 + " hours, " + m1 + " minutes to t1");

        t1.add(h1, m1);

        System.out.println("New t1 state: " + t1);

        // creating another TimeSpan object, testing compareTo method using the

        // two objects

        TimeSpan t2 = new TimeSpan(10, 20);

        System.out.println("New object t2: " + t2);

        System.out.println("t1.compareTo(t2): " + t1.compareTo(t2));

        System.out.println("t2.compareTo(t1): " + t2.compareTo(t1));

        System.out.println("t1.compareTo(t1): " + t1.compareTo(t1));

   }

}

/*OUTPUT*/

New object t1: 13h30m

Adding 3 hours, 40 minutes to t1

New t1 state: 17h10m

New object t2: 10h20m

t1.compareTo(t2): 1

t2.compareTo(t1): -1

t1.compareTo(t1): 0

4 0
2 years ago
I am doing keyboarding keyboarding is very boring and yeah
Bingel [31]

Answer:

yes I agree with you

Explanation:

I have been keyboarding for the past 7 hours.

7 0
2 years ago
Read 2 more answers
Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

               = 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 0
2 years ago
If byte stuffing is used to transmit Data, what is the byte sequence of the frame (including framing characters)? Format answer
Lerok [7]

Answer:

Correct Answers: 01h 79h 1Bh 78h 78h 1Bh 7Ah 04

Explanation:

Solution is attached below

4 0
2 years ago
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