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Delicious77 [7]
3 years ago
7

When configuring an adsl installation where should you to install the dsl filters?

Computers and Technology
1 answer:
xxTIMURxx [149]3 years ago
5 0

<span>Filter installation in ADSL setup, the DSL filters should on connections leading to an analog phone. The DSL modem and phone were share on the same jack. ADSL is a type of digital subscriber line machinery, a data communication that enables faster data transfer over copper telephone lines.</span>

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Monica is teaching herself to paint. What web resource would be the most helpful
strojnjashka [21]

Answer:

Video Tutorial

Explanation:

The others don’t make sense.

3 0
3 years ago
An additional factor in how an element is rendered is that properties are passed from a parent element to its children in a proc
ValentinkaMS [17]

Answer: Style inheritance

Explanation:

Style inheritance is used to design style sheets. It is a process in which properties are inherited by children from a parent element. For example you wish that all text on a page use same font color i.e. red. You can apply they following style for <body> tag like {body color:red;}. All the elements in the web page will inherit this font color. This is better to use than to create styles for each tag. Every heading and paragraph will be displayed in this font color until you define different color for particular element.

3 0
3 years ago
How would you write this using Java: Use a TextField's setText method to set value 0 or 1 as a string?
katrin [286]

Answer:

Explanation:

The following Java code uses JavaFX to create a canvas in order to fit the 10x10 matrix and then automatically and randomly populates it with 0's and 1's using the setText method to set the values as a String

import javafx.application.Application;

import javafx.scene.Scene;

import javafx.scene.control.TextField;

import javafx.scene.layout.GridPane;

import javafx.stage.Stage;

public class Main extends Application {

   private static final int canvasHEIGHT = 300;

   private static final int canvasWIDTH = 300;

   public void start(Stage primaryStage) {

       GridPane pane = new GridPane();

       for (int i = 0; i < 10; i++) {

           for (int j = 0; j < 10; j++) {

               TextField text = new TextField();

               text.setText(Integer.toString((int)(Math.random() * 2)));

               text.setMinWidth(canvasWIDTH / 10.0);

               text.setMaxWidth(canvasWIDTH / 10.0);

               text.setMinHeight(canvasHEIGHT / 10.0);

               text.setMaxHeight(canvasHEIGHT / 10.0);

               pane.add(text, j, i);

           }

       }

       Scene scene = new Scene(pane, canvasWIDTH, canvasHEIGHT);

       primaryStage.setScene(scene);

       primaryStage.setMinWidth(canvasWIDTH);

       primaryStage.setMinHeight(canvasHEIGHT);

       primaryStage.setTitle("10 by 10 matrix");

       primaryStage.show();

   }

   public static void main(String[] args) {

       Application.launch(args);

   }

}

3 0
3 years ago
identify three ways we use analog and digital signals in our everyday lives. Describe how radio telescopes are used to explore s
spayn [35]
We use analog and digital signals in our everyday lives with the radio, the cell phone and with our own computers. our computers and our cell phones are all digital now but radio's are still considered analog. Radio telescopes explore space by using radio waves and signals to almost like scan space and collect data like a sonar in a submarine. 
8 0
3 years ago
Read 2 more answers
A detective agency is looking to bust an arms sale. According to available intelligence, the sale is likely to happen in one of
Paha777 [63]

Answer:

Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}

H(X)=5.4224 bits per symb

H(X|Y="not C")=0.54902 bits per symb

Explanation:

P(B)=2P(C)

P(A)=2P(B)

But

P(A)+P(B)+P(C)=1

4P(C)+2P(C)+P(C)=1

P(C)=1/7

Then

P(A)=4/7

P(B)=2/7

Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}

iii

If X={A,B,C}

and P(Xi)={4/7,2/7,1/7}

where  Id =logarithm to base  2

Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}

=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}

=5.4224 bits  per symb

if P(C)  =0

P(A)=2P(B)

P(B)=1/3

P(A)=2/3

H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)

=0.54902 bits per symb

4 0
3 years ago
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