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nirvana33 [79]
2 years ago
13

Provide the definition of an abstract class named DesktopComponent that contains the following:

Computers and Technology
1 answer:
Bess [88]2 years ago
7 0

package mypackage; // Whatever package this should be in.

public abstract class DesktopComponent {

   

   private String type;

   // Alternatively you may want a final variable.

   // private final String type;

   

   public DesktopComponent(String type)

   {

       this.type = type;

   }

   

   abstract void onClick();

   

   

 

}

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What are the three general methods for delivering content from a server to a client across a network
Tju [1.3M]

Answer:

Answered below.

Explanation:

The three general methods consist of unicasting, broadcasting and multicasting.

Casting implies the transfer of data from one computer (sender) to another (recipient).

Unicasting is the transfer of data from a single sender to a single recipient.

Broadcasting deals with the transfer of data from one sender to many recipients.

Multicasting defines the transfer of data from more than one sender to more than one recipients.

8 0
2 years ago
How many answer or point do I answer/need to sent message?
Grace [21]

I'm a little confused about the question, but if you're asking how many points you need to ask a question, the minimum is 10.


4 0
3 years ago
The publisher tab in the application control allows you to manage the various certificates that are used to do what to binaries?
artcher [175]

The publisher tab in the application control allows you to manage the various certificates that are used to sign to binaries. Publisher allow executables of a particular vendor, signed by a security certificate issued to the vendor by a Certificate Authority. Also, all applications and binary files either added to or modified on an endpoint that are signed by the certificate are automatically added to the whitelist.

3 0
2 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
List the six external parts or "peripherals" of a computer system and identify which are output and which are input devices/ Lis
vitfil [10]
Keyboard-Input
Mouse-Input
<u></u>Monitor-Output
Speakers-Output
Printer-Output
Hard Drive<span>-Output</span>
3 0
3 years ago
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