An operating system is an example of _______. The hard drive, keyboard and monitor are example of_______.​

Answer:

A private IP address should be implemented but a PAT and a static NAT should be configured for the computers and the intermediate devices respectively.

Explanation:

There are two types of IP addresses, they are public and private addresses. The public address is an IP4 address routable on the internet while the private address is not routable.

Internet service providers do charge for the use of public IP addresses they render, which can be very costly. To reduce the company's expenses on the internet, the company is networked with private addresses and are mapped for network address translation ( NAT) which assigns a public global address to devices accessing the internet.

Answer:

is to sit up straight

Explanation:

more blood flows through ur head and u can procecs more information and be more active on your work

Answer: Dark background and light text.

Explanation:

Assuming when you mean by "light text", you mean by white text. This should be able to be readable by users visiting the site.

Answer:

Contrasting background and text

Explanation:

When selecting a color scheme for a website, designers often choose complementary colors that keep the visitor on the website.

Your web page should have three main colors: one for the background, one for the text, and one for the headings. The colors you select should complement each other and be on opposite sides of the color wheel.

Also i got it right on my test!

Answer: The one that gives you more vibe that its correct.

Explanation:

The one that you feel most strongly is accurate is the  best describes the overall structure of the passage.

A passage's internal organizational pattern is defined by its text structure. The author's intent behind the text is supported by the text's structure. When determining the structure of a document, transitions—words or phrases that demonstrate how things are related—can be highly beneficial.

These five text structures—description, sequence, cause and effect, compare and contrast, and issue and solution—are taught in this lesson because they are frequently utilized in informational and nonfiction texts.

The table of contents, the index, headings, captions, bold words, illustrations, pictures, the glossary, labels, graphs, charts, and diagrams are some of the most frequent text elements in books.

Thus, he one that you feel most strongly is accurate.

For more information about structure of the passage, click here

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Answer:

(a) Variable

Explanation:

In the programming world, a variable is used to store data that can be referenced and modified in a computer program. It typically refers to the label for the location (in memory) of a particular data used in programs. For example;

var b = 12.

This implies that the value 12 is stored in a location in memory that is labeled as b. Anytime the value 12 is needed either for retrieval or modification purposes, it can be referenced by simply calling the variable b.

Answer

Answer:

WAHTSTFYHHTHIEBWIBWDKF

Explanation:thx

FOR A SEC

Answer:

SYN and ACK.

Explanation:

In Computer Networking, a control bit flag is used in transmission control protocol (TCP) to denote a specific state of connection and typically indicates a bit of information (1 bit).

The control bit flags used during the three-way handshake are SYN and ACK. A three-way handshake is used in transmission control protocol (TCP) to establish a reliable connection.

The three-way handshake refers to a client-server communication model which happens in three steps or stages;

1. SYN: this is where the client establishes a connection with the server.

2. SYN-ACK: the server then immediately responds to the server by acknowledging the request.

3. ACK: here the client acknowledges the response of the server and thus create a reliable connection.

Answer:

Chart area

i hope this helps

Answer:

Named Constant

Explanation:

blank is a memory location whose value cannot change during the execution of the program is Named Constant

Answer: D, Throughput

Explanation: Throughput is a measure of how many units of information a system can process in a given amount of time.

Answer:

No, you can't.

Explanation:

SSH is an acronym for Secure Socket Shell and is a networking protocol used for providing remote authentication through a public-key cryptography over an unsecured network such as the internet.

Hence, it is a network protocol that is commonly implemented by using a client-server model; where a computer acts as a SSH client while the other network device acts as the host or SSH server.

When using SSH to remotely access a Cisco router, you cannot see the terminal password because it is Linux based and encrypted through the use of symmetrical encryption, asymmetrical encryption or hashing

Answer:

Here is the C function that counts the letters in a string:

int LetterCount(char string[]){  //function to count letters in a string passed as parameter

 string[100];  // char type string with size 100

  int i, letters;  // letter variable stores the count for no. of letters in string

   i = letters = 0;  //initialize variables i and letters to 0

  while (string[i] != '\0')  { // loop iterates through the entire string until end of string is reached

    if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  { // if condition checks the letters in the string

     letters++;    }  // increments 1 to the count of letters variable each time a letter is found in the string

    i++;  }  //increments value of i to move one character forward in string

   printf("Number of Letters in this String = %d", letters);   // displays the number of letters in the string

   return 0; }                              

Explanation:

Here the question means that the function should count the letters in a string. So the letters are the alphabets from a to z and A to Z.

Here is the complete program:

#include <stdio.h>   // to use input output functions

int LetterCount(char string[]){  // method that takes a character string as parameter and counts the letters in the string

string[100];  

int i, letters;

i = letters = 0;

while (string[i] != '\0')   {

 if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z'))

   {letters++;  }

   i++; }

   printf("Number of Alphabets in this String = %d", letters);  

   return 0;}  

int main(){  // start of main function

  char string[100];  //declares a char array of string

   printf("Please Enter a String : ");   //prompts user to enter a string

  fgets(string,100,stdin);  //get the input string from user

   LetterCount(string); } // calls method to count letters in input string

I will explain this function with an example

Lets suppose the user input "abc3" string

string[100] = "abc3"

Now the function has a while loop that while (string[i] != '\0')  that checks if string character at i-th position is not '\0' which represents the end of the character string. As value of i = 0 so this means i is positioned at the first character of array i.e. 'a'

At first iteration:

i = 0

letter = 0

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )   if condition checks if the character at i-th position of string is a letter. As the character at 0-th position of string is 'a' which is a letter so this condition evaluates to true. So the statement letter++ inside if condition executes which increments the letter variable to 1. So the value of letter becomes 1. Next statement i++ increments the value of i to 1. So i becomes 1. Hence:

i = 1

letter = 1

At second iteration:

i = 1

letter = 1

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 1st position of string is 'b' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 2

letter = 2

At third iteration:

i = 2

letter = 2

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 2nd position of string is 'c' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 3

letter = 3

At fourth iteration:

i = 3

letter = 3

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 3rd position of string is '3' which is not a letter but a digit so this condition evaluates to false. So the statement letter++ inside if condition does not execute. Now i++ executes which increments this variable to 1. Hence:

i = 4

letter = 3

Now the loop breaks because while (string[i] != '\0') condition valuates to false as it reaches the end of the string.

So the statement: printf("\n Number of Letters in this String = %d", letters); executes which prints the value of letters on the output screen. Hence the output is:

Number of Letters in this String = 3

Answer:

go to your dashboard and press the number under grade to date (your percentage in the class) and it should show all your assignments that you have done and that still need to be done

Answer:

Here is the recursive method NoOfOnes :

public static int NoOfOnes(int N)    { // method that takes number N as argument and returns the number of 1’s in the binary representation of N

       if(N==0)  { //base case 1

           return 0;        } //if the value of N is equal to 0 then method returns 0

       else if(N==1){ //base case 2

           return 1;        } // if the value of N is equal to 1 then method returns 1

       else if(N%2==1)  { //if N is odd (recursive case)

           return NoOfOnes(N/2)+1;   } //calls method recursively and divides N by 2 plus 1

       else{ //when N i even (recursive case)

           return NoOfOnes(N/2);   }   }   //calls method recursively and divides N by 2  to return number of 1's in binary representation of N

Explanation:

Here is the complete JAVA program:

import java.util.Scanner; //to take input from user

public class Main {

   public static int NoOfOnes(int N)    {//the method that takes number N as argument and returns the number of 1’s in the binary representation of N

       if(N==0)  { //base case 1

           return 0;        }

       else if(N==1){//base case 2

           return 1;        }

       else if(N%2==1)  {//recursive case when N is odd

           return NoOfOnes(N/2)+1;   }

       else{ //recursive case when N is even

           return NoOfOnes(N/2);   }   }    

public static void main(String[] args) { //start of main function

       Scanner scan = new Scanner(System.in); //creates object of Scanner

       System.out.print("Enter n: ");//prompts user to enter value of N

       int n = scan.nextInt(); //reads the value of n from use

 System.out.print("number of 1’s in the binary representation of " +n+" is: "+NoOfOnes(n)); } } //displays the number of 1’s in the binary representation of n

i will explain the working of the function with an example

Lets say N = 6

if(N==0) this statement evaluates to false as N is not 0

else if(N==1 this statement also evaluates to false as N is not 1

else if(N%2==1) this statement also evaluates to false because N is not odd. If we take modulo of 6 by 2 then the remainder is 0 because 6 is completely divisible by 2 so N = 6 is an even number. So the else part is executed:  

       else{

           return NoOfOnes(N/2);}

This statement calls NoOfOnes recursive by passing N/2 to the method. So this becomes:

return NoOfOnes(6/2)

6/2 = 3

NoOfOnes(3)

Now this method is called again with the value of N = 3. Since 3 is an odd number so the recursive case of odd N is executed:

else if(N%2==1)  {

           return NoOfOnes(N/2)+1;   }

NoOfOnes(3/2)+ 1

NoOfOnes(1) + 1

Now the method is called again with value of N=1. The base case 2 is executed because N==1 which returns 1

else if(N==1){

           return 1;   }

So this becomes

NoOfOnes(1) + 1

1 + 1

= 2

So the method return 2 as the number of 1’s in the binary representation of N = 6

Hence the output of the above program with N = 6 is

number of 1’s in the binary representation of 6 is: 2          

Answer:

a. True

Explanation:

Objects suffer inertia before a transition is achieved.  Transition is the movement of an object from one state to another.  But, this transition does not happen on its own.  It is usually triggered by an event or series of events.  This means that when the object moves from its prior state to a subsequent state, an event must have occurred to trigger the transition.  Objects, therefore, depend on events or an outside force to remove their inertia.  They are usually in a state of inertia until an outside stimulus stimulates them with some momentum.

Answer:

Here is the recursive function:

bool isPalindrome(string& MyString,int start, int end){    // function that takes as argument a string and two arguments that are bounds indices

       while(start < MyString.length() && !isalpha(MyString[start])){  //while loop continues to execute as long as start does not exceed the length of the string AND does not point to a alphabetic letter

           start++;         } // increments start by 1 moving it to point the next  character

       while(end >= 0 && !isalpha(MyString[end])){  //while loop continues to execute as long as end does not gets less than 0 AND does not point to a alphabetic letter

           end--;         }    // decrements end by 1 moving it to point the next  character backwards

       if(start > end){  //the base case

           return true;        

       } else {    //recursive case

           return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1));         }    }  //The recursive case is to AND the result of comparing the  alphabetic letters at start and end with result of the function  (called recursively) after without both of the letters

Explanation:

Here is the complete C++ program:

#include <iostream>  //to use input output functions

#include <cctype>  //functions to manipulate characters

using namespace std; //to identify objects as cin cout

bool isPalindrome(string&,int start, int end);  // function prototype

int main(){  //start of main function

       string MyStrings;        //declare a string type variable to hold string of characters

       cout<<"Enter a string : ";  //prompts user to enter a string

       getline(cin,MyStrings );  //gets and reads the string from user

       while(MyStrings != "quit"){        // continues to take input from user until user enters quit

       if(isPalindrome(MyStrings ,0,MyStrings.length()-1)== true){  //calls function that checks if the string is a palindrome

       cout<<MyStrings<<" is a Palindrome\n";}  //if the input string is a palindrome then display this message

       else  

       {cout<<MyStrings<<" is not a Palindrome\n";}  //if the input string is not a palindrome then displays this message

       cout<<"Enter a string : ";  //prompts user to enter a string

       getline(cin,MyStrings); }    }        //reads a string from user

   bool isPalindrome(string& MyString,int start, int end){  //function definition        

       while(start < MyString.length() && !isalpha(MyString[start])){  

           start++;         }          

       while(end >= 0 && !isalpha(MyString[end])){

           end--;         }        

       if(start > end){

           return true;        

       } else {              

           return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1));         }    }

I will explain the working of the function:

First the start variable is pointed to the start of the MyString and the end variable points to the last character of MyString    

while(start < MyString.length() && !isalpha(MyString[start]))

The above statement has a while loop which continues to execute as long as start does not exceed the string length and  it does not point to a letter. It has a AND logical operator that means both the above condition have to be true in order to continue this while loop. Here isalpha() method is used that checks if the letter/character of MyString is an alphabet or not. At each iteration start is incremented to 1 to point the next  character of MyString.

while(end >= 0 && !isalpha(MyString[end]))

The above statement has a while loop which continues to execute as long as end does not becomes less than 0 and  it does not point to a letter, Here isalpha() method is used that checks if the letter/character of MyString is an alphabet or not. At each iteration end is decremented to 1 to point the next  character of MyString in backward or in reverse order.

if(start > end){  

The above statement is a base case which is the IF condition that checks if the start is greater than end. This basically shows that there are no more alphabetic characters and this base case returns true as it reads the same from both sides.

else {              

           return ( ( toupper(MyString[start]) == toupper(MyString[end]) ) && isPalindrome(MyString,start+1,end-1));         }    }

The above statement is a recursive part. It uses AND logical operator and compares the characters at start and end with the result of the function  after excluding both of the characters pointed by start and end.

The program along with its output is attached.

Answer:

Here is the JAVA application that determines which if any of the following files are stored in folder where you have saved the exercises created.

import java.nio.file.*;   // used to manipulate files and directories

import java.nio.file.attribute.*;  //used to provide access to file and file attributes

import static java.nio.file.AccessMode.*;  // provides access modes to test the accessibility of a file

import java.io.IOException;  //for exception handling

class FindSelectedFiles {

public static void main(String[] args) { //start of main function

   Path f1 = Paths.get("/root/sandbox/autoexec.bat"); //creates Path object f1 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f2 = Paths.get("/root/sandbox/CompareFolders.java");  //creates Path object f2 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f3 = Paths.get("/root/sandbox/FileStatistics.class");   //creates Path object f3 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f4 = Paths.get("/root/sandbox/Hello.doc"); //creates Path object f4 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   int count=0; //used to count the number of files found  

try{  // used to define a code chunk to be tested for errors

 f1 = Paths.get("/root/sandbox/autoexec.bat");  // Converts a path string when joined form a path string to a Path

 System.out.println(f1);  //prints the file

     /* getFileSystem method is used to return the file system that created this Path object f1 and checkAccess Verifies access and throws exception if something is wrong

 f1.getFileSystem().provider().checkAccess(f1);  

 System.out.println(f1 +" File Exists\n");  //if file exists then displays this message

 count++;  }  //adds 1 to the count of number of files found

 catch (IOException e) {  //a code chunk to be executed if error occurs in the try part

  System.out.println("File Doesn't Exist\n"); }  //if file not found displays this message

try{  //works same as above try block but for 2nd folder and object f2

 f2 = Paths.get("/root/sandbox/CompareFolders.java");

 System.out.println(f2);

 f2.getFileSystem().provider().checkAccess(f2);

 System.out.println(f2 +" File Exists\n");

 count++;  }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

try{  //works same as above try block but for 3rd folder and object f3

 f3 = Paths.get("/root/sandbox/FileStatistics.class");

 System.out.println(f3);

 f3.getFileSystem().provider().checkAccess(f3);

 System.out.println("File Exists\n");

 count++; }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

try{  //works same as above try block but for 4th folder and object f4

    f4 = Paths.get("/root/sandbox/Hello.doc");

    System.out.println(f4);

 f4.getFileSystem().provider().checkAccess(f4);

 System.out.println(f4 +" File Exists\n");

 count++; }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

   System.out.println(count+" of the files exist");  }  } //displays the number of files found

Explanation:

If you want to directly pass the file name as argument to Paths.get() method then you can use the following code:

   Path f1 = Paths.get("autoexec.bat");

   Path f2 = Paths.get("CompareFolders.java");

   Path f3 = Paths.get("FileStatistics.class");

   Path f4 = Paths.get("Hello.doc");

The program determines if any, of the files from autoexec.bat, CompareFolders.java, FileStatistics.class, and Hello.doc. are stored in the folder. If a file is found then the program displays the message File exists and add 1 to the count variable each time a file is found. If all the files are found then the program returns the count as 4 otherwise the program display the number of files found by displaying the computed final value of count in the output. Lets say all the files are found then the output of the above program is:

autoexec.bat                                                                                                                                                   autoexec.bat File Exists  

                                                                                                                                                      CompareFolders.java                                                                                                                                            CompareFolders.java File Exists                                                      

                                                                                                                                                         FileStatistics.class                                                                                                                                           File Exists                                                                                        

                                                                                                                                                     Hello.doc                                                                                                                                                      Hello.doc File Exists  

4 of the files exist    

Answer:

public CarRecord(){

       yearMade = 0;

       vehicleIdNum = -1;

   }

Explanation:

A constructor is used to initialize an object instantly. It does not have a return type and has the same name as class.

Create a constructor and set the year as 0 and vehicleIdNum as -1.

This constructor is the default constructor, initializes the default values for the variables, for this class.

A constructor is a block of code that is called and executed each time an object is created.

The constructor in Java is as follows:

public CarRecord(){

       yearMade = 0;

       vehicleIdNum = -1;

  }

A constructor would have the same name as the class.

The class of the program is CarRecord; so the constructor name is CarRecord()

From the question, variables year and vehicles ID are to be initialized to 0 and  -1 respectively.

So, the variable initialization is:

       yearMade = 0;

       vehicleIdNum = -1;

Hence, the complete code segment is:

public CarRecord(){

       yearMade = 0;        vehicleIdNum = -1;

  }

Read more about constructors at:

https://brainly.com/question/14608885

Answer:

Six factors that influence one's design choices in data visualization are;

1. subject-matter appeal

2. personal taste

3. attitude and emotion

4. dynamic of need

5. what they need to know

6. subject-matter knowledge.

Explanation:

The subject-matter appeal is the interest the viewers show to the data presented to them, if it really relates to their interest.

Personal taste is the theme from color to the graphic tool used to display the data. Dynamics of need depicts the viewers' choice to know the data they are presented with.

Subject-matter knowledge is the prior knowledge of the data presented, do they have prior knowledge or not.

What they need to know is a factor that helps to answer the viewers' question, does this have the knowledge I seek?

Answer:

I think is C but I'm not completely sure

Answer:

all of them except avoiding setting goals

Explanation:

Part of personal growth increasing self-awareness, improving quality of life, and developing skills. The correct options are a, b, and d.

Understanding yourself and pushing yourself to realize your full potential are both parts of the process of personal growth. It entails continuously considering your identity and your plans for getting there.

Utilize your time management abilities more effectively, and take better care of yourself. Stop putting things off. Set boundaries for your use of social media. Increase the amount of self-care activities you do each day.

The main components of personal growth are increasing awareness of oneself, enhancing living quality, and increasing skills of yourself.

Therefore, the correct options are:

a. Increasing self-awareness.

b. improving quality of life

d. developing skills

To learn more about personal growth, visit here:

https://brainly.com/question/28146493

#SPJ2

The question is incomplete. Your most probably complete question is given below:

Increasing self-awareness

improving quality of life

avoiding setting goals

developing skills

Answer:

import csv

def ideal_weight( filename ):

   with open( "filename", "r" ) as file:

       for row in csv.DictReader( file):

           print ( "Name of person: ",row["name"] )

           person_weight = ( 110 * row["height"] )/ 5

           print ( "Ideal weight: ", person_weight )

Explanation:

The python source code above is a function which recieves a csv file argument, opens the file and reads each row an ordered dictionary. The function prints the name of each person in the file and their ideal weight.

Answer:

1)Numbers

2)Text

3) Graphics

Explanation:

three kinds of information stored in a computer are

1)Numbers

2)Text

3)Graphics

Answer:

d. left-aligned

Explanation:

They are accustomed to seeing paragraphs that are left aligned. The left alignment is the most common type of alignment. This type of text format aligns text at the left-hand side of a page or paper. In areas areas or countries were english is spoken like the United States, text editors are usually in left alignment. The reason for this is because they read from left to right in these countries.

Answer: I would say D. because it is highlighted text draws the attention of the reader

Explanation:

Answer:

D. highlight the text

Explanation:

if your were trying to grab a readers attention highlighting the text is the most noticeble.

Answer:

Mirror

Explanation:

No explanation

Answer:

I think your answer is Light. Its not silver because thats

just what the camera is made of. Shutters I don't know what it does. Im positive that it is Light!