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Strike441 [17]
4 years ago
13

Sofia is reading a difficult text for class and worries that she won't complete it by the given deadline. How can a text-

Computers and Technology
2 answers:
castortr0y [4]4 years ago
5 0

Answer:

It can increase the pace of her reading

Explanation:

alex41 [277]4 years ago
5 0

It Can increase the pace of her reading

Hi༼ つ ◕_◕ ༽つ

You might be interested in
What is the term that is used to describe a computer system that could store literary documents, link them according to logical
madreJ [45]

Answer:

hypertext

Explanation:

Based on the information provided within the question it can be said that the type of computer system function being described is known as hypertext. This displays text to the computer display that references other literary documents for immediate access to them. These are documents with a specific relation to the text that is being displayed and also allows readers to comment and annotate what they read on the documents.

5 0
3 years ago
create a new Java application called "CheckString" (without the quotation marks) according to the following guidelines.** Each m
jok3333 [9.3K]

Answer:

The Java code is given below

Explanation:

import java.io.*;

import java.util.Scanner;

import java.util.ArrayList;

public class StringCheck{

//There are several implementation approaches. This is one example framework/outline which might be helpful. Please feel free to try other approaches.

public static void checkWord(String word) throws Exception {

// Uses charAt method to test if the first letter of string variable word

// is a character. If not, throw new exception

if(Character.isLetter(word.charAt(0))){

 return;

}else{

 throw new Exception("This is not a word");

}

}

public static String getWord() {

// Declare a local scanner

// Prompt the user to enter a word

// Think about using a loop to give the user multiple opportunities to correctly enter a string

// Read into a string

// Call checkWord method passing the string as a parameter

// checkWord can throw an exception; call checkWord in a try/catch block

// Return the string to main if a valid string

Scanner sc = new Scanner(System.in);

boolean st = true;

String word = null;

while(st){

 System.out.println("Enter a Word : ");

 word = sc.next();

 try{

  checkWord(word);

  st = false;

 }catch(Exception e){

  System.out.println(e);

  st = true;

 }

}

sc.close();

return word;

}

public static void writeFile(String[] arrayToWrite, String filename) throws IOException {

// Example using FileWriter but PrintWriter could be used instead

// Create a FileWriter object

FileWriter fileWordStream = new FileWriter(filename);

// Use a loop to write string elements in arrayToWrite to fileWordStream

for(int i = 0;i<arrayToWrite.length;i++){

fileWordStream.write(arrayToWrite[i]+System.lineSeparator());

}

fileWordStream.flush();

fileWordStream.close();

// In the loop use the lineSeparator method to put each string on its own line

}

public static ArrayList readFile(String filename) throws FileNotFoundException, IOException {

// Declare local ArrayList

ArrayList<String> words = new ArrayList<>();

// Create a new File object using filename parameter

File file = new File(filename);

// Check if the file exists, if not throw a new exception

if(!file.exists()){

 throw new FileNotFoundException("File not Found");

}

// Create a new BufferedReader object      

Scanner fsc = new Scanner(file);

// use a loop and the readLine method to read each string (on its own line) from the file

while(fsc.hasNextLine()){

words.add(fsc.nextLine());

}

fsc.close();

// return the filled ArrayList to main

return words;

}

public static void main(String[] args) {

// create a string with literal values to write to the file

String[] testData = {"cat", "dog", "rabbit"};

// Create an ArrayList for reading the file

ArrayList<String> words = new ArrayList<>();

// Declare a string variable containing the file name "data.txt"

String file = "data.txt";

// Call getWord, assign the returned string to a variable and display it

String word = getWord();

System.out.println("The word is : "+word);

// Call writeFile and readFile methods in a try block

try{

writeFile(testData,file);

words = readFile(file);

// Printout the contents of the ArrayList after the call to readFile

for(int i = 0;i<words.size();i++){

 System.out.println(words.get(i));

}

}catch(FileNotFoundException e){

System.out.println(e);

}catch(IOException e){

System.out.println("IOException: error occured in input output");

}

// catch two types of exceptions

}

}

4 0
3 years ago
This morning when Paul turned on his computer at work, it would not boot. Instead, Paul reported that he heard a loud clicking n
IRISSAK [1]

Answer:

The problem with Paul's computer is that the magnetic hard drive failed.

Explanation:

A magnetic hard drive or hard disk is a non-volatile storage device, which is used to store different information on a computer.

The magnetic disk is made up of numerous magnetic plates, these plates are divided into sectors and tracks. A spindle is present in the middle to rotate the combined plates or the whole unit.

When there is a clicking noise coming from the magnetic disk, it could mean that one of the plates is damaged or if the power to the computer is inconsistent. This can greatly affect the performance of your computer.

3 0
4 years ago
What is a client server network and its features?​
maks197457 [2]

Answer:

A client-server network is the medium through which clients access resources and services from a central computer, via either a local area network (LAN) or a wide-area network (WAN), such as the Internet. ... A major advantage of the client-server network is the central management of applications and data.

8 0
3 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

5 0
3 years ago
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