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snow_lady [41]
3 years ago
7

The unique global address for a web page or other resource on the internet is called the ________.

Computers and Technology
1 answer:
Mila [183]3 years ago
8 0
URL or Uniform Resource Locator
You might be interested in
Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0
3 years ago
How to get an object from a container in java.
Vinil7 [7]
If it stores any type of objects mixed, use Object as storage class. All classes inherit from Object and for primitives use their respective wrapper classes. Or just use one of the bazillion container classes that already exist.
3 0
3 years ago
You recently created several new user accounts in the Sales OU and configured them with the appropriate group membership, logon
irga5000 [103]

Answer:

Disable the account until the employee is ready to use it.

Explanation:

6 0
3 years ago
Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volu
marysya [2.9K]
<h2>Question:</h2>

Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volume of a pyramid with a rectangular base. Relevant geometry equations:

Volume = base area x height x 1/3

Base area = base length x base width.

(Watch out for integer division)

import java.util.Scanner;

public class CalcPyramidVolume {

   /* Your solution goes here */

   public static void main (String [] args) {

     System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0,   1.0));

     return;

    }

}

<h2>Answer:</h2><h2></h2>

import java.util.Scanner;

public class CalcPyramidVolume {

   /* Your solution goes here */

   public static void main (String [] args) {

       System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0, 1.0));

       return;

   }

   

   //Begin method definition

   public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){

       

       //First, calculate the base area of the pyramid

       //store the result in a double variable

       double baseArea = baseLength * baseWidth;

       

       //Then, calculate the volume of the pyramid

       //using the base area and the base width

       double volume = 1 / 3.0 * baseArea * pyramidHeight;

       

       //return the volume

      return volume;

   }

}

<h2>Output:</h2>

Volume for 1.0, 1.0, 1.0 is: 0.3333333333333333

<h2>Explanation:</h2>

The code above contains comments explaining the important lines of the code. The output of the code has also been provided above.

The parts of the code that are worth noting are:

(i) The return type of the method pyramidVolume should be a <em>double</em> since calculations are done using <em>double</em> values. i.e the method header should be written as

<em>public static double pyramidVolume(){</em>

<em>}</em>

<em />

(ii) The method requires three(3) parameters of type double: baseLength, baseWidth and pyramidHeight. These should be included in the method. Therefore, the complete header definition should be:

<em>public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){</em>

<em />

<em>}</em>

(iii) The formula for calculating the volume of the pyramid should be:

volume = 1 / 3.0 * baseArea * baseWidth;

Rather than;

volume = 1 / 3 * baseArea * baseWidth;

This is because integer division yields integer result. If 1 / 3 is evaluated, the result will be 0 since the decimal part will be truncated thereby making the result of the volume = 1 / 3 * baseArea * baseWidth will be 0.

Therefore, the work around for that should be to write 1 / 3.0 or 1.0 / 3.0 or 1.0 / 3.

(iv) After the calculation, the result of volume should be returned by the method. This will enable a proper call by the main method for execution.

7 0
3 years ago
Entering the formula =sum(c5:c18) in cell c19 will result in which of the following?
antiseptic1488 [7]
If you mean Excel formula sum(c5:c18) in cell c19 will show <span>c. the total of cells c5 to c18 will appear in cell c19. It's simple operation that simplyfies counting different values.</span>
5 0
3 years ago
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