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Scrat [10]
2 years ago
7

Write a MATLAB function named lin_spaced_vector with two inputs and one return value. The first input will be a single real numb

er representing a lower bound The second input will be a single real number representing an upper bound The return value must be a list of 200 numbers evenly spaced between the lower bound and the upper bound.

Computers and Technology
1 answer:
ycow [4]2 years ago
4 0

Explanation:

==================  

lin_spaced_vector.m  

==================  

function out=lin_spaced_vector(in1,in2)%defining function

out=linspace(in1,in2,200);%200 spaced numbers between in1 and in2

end​

===================  

Executable File

===================

clear all%clears history

clc%clears screen

lin_spaced_vector(1,10)%calling function​

clear all

clc

lin_spaced_vector(1,10)

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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
Edhesive unit 2 lesson 5 coding activity 1 Write code which creates three regular polygons with 11, 14 and 19 sides respectively
SVEN [57.7K]

Answer:

public class Polygon {

   private String name;

   private int sides;

   private double sideLength;

   public Polygon(String name, int sides, double sideLength) {

       if (sideLength <= 0) throw new IllegalArgumentException("Length cannot be zero or negative.");

       if (sides <= 0) throw new IllegalArgumentException("Sides cannot be zero or negative.");

       this.name = name;

       this.sides = sides;

       this.sideLength = sideLength;

   }

   public String getName() {

       return name;

   }

   public void setName(String name) {

       this.name = name;

   }

   public double getSideLength() {

       return sideLength;

   }

   public void setSideLength(double sideLength) {

       if (sideLength <= 0) throw new IllegalArgumentException("Length cannot be zero or negative.");

       this.sideLength = sideLength;

   }

   public int getSides() {

       return sides;

   }

   public void setSides(int sides) {

       this.sides = sides;

   }

   (use the at sign here)Override

   public String toString() {

       return "regular " + name + " with side length " + String.format("%.1f", sideLength);

   }

}

public class TestPolygon {

   public static void main(String[] args) {

       Polygon sides11 = new Polygon("hendecagon", 11, 1);

       Polygon sides14 = new Polygon("tetrakaidecagon", 14, 1);

       Polygon sides19 = new Polygon("enneadecagon", 19, 1);

       System. out. println(sides11);

       System. out. println(sides14);

       System. out. println(sides19);

   }

}

Explanation:

This java source code defines a class that creates a regular polygon based on the number of sides given to it.

Below is a screenshot of the program code and output.

4 0
1 year ago
Write a static generic method PairUtil.minmax that computes the minimum and maximum elements of an array of type T and returns a
Gnesinka [82]

Answer:

Explanation:

The following code is written in Java. It is hard to fully create the code without the rest of the needed code including the T class and the Measurable interface. Regardless the following code can be implemented if you have that code available.

 public static T minmax(ArrayList<T> mylist) {

       T min = new T();

       T max = new T();

       for (int x = 0; x < mylist.size(); x++) {

           if (mylist.get(x) > max) {

               max = mylist.get(x);

           } else if (mylist.get(x) < min) {

               min = mylist.get(x);

           }

       }

       

       return (min, max);

   }

5 0
2 years ago
A process is moved to wait queue when I/O request is made with:_______ A. non-blocking I/O B. blocking I/O C. asynchronous I/O D
Sidana [21]

Answer:

B. blocking I/O

Explanation:

           Most of the input and output request placed  to the computer considers the blocking request. It means that the controls given cannot be returned to the given application until and unless the input/output is complete.

           Thus, blocking the input/output does not return till the input and output is complete.

            With a blocking I/O, the process is moved to a wait queue when the I/O request is made, and it moved backs to the ready queue as soon as the request is complete, thereby allowing the other processes to run in the meantime.

Thus (B) blocking I/O is the answer.

7 0
2 years ago
Susie works for an architectural firm and the partners have always drawn the plans for projects by hand. Though Susie learned ho
lidiya [134]

Answer:

The correct answer to the following question will be CAD.

Explanation:

CAD - It stands for Computer-aided Design. CAD is a software used for creating some 3D designs, creative drawing or some technical illustrations and is basically used by the engineers and artists.

Many projects based on electronic can be designed with the help of this software, electronic parts like circuit boards, etc.

Various types of CAD are as follows :

  • 2D (Two dimensional)
  • 3D (Three dimensional)
  • 3D wire frame
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This enables efficiency in design's quality, also increase in engineer and artist's productivity. So, many of the computer programs should recommend the Computer-aided design software to there project partners for some 3D and effective look on their project.

3 0
2 years ago
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