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Sloan [31]
3 years ago
10

Do you like PC? Why? (I need your opinion for research)

Computers and Technology
2 answers:
Murljashka [212]3 years ago
5 0
I like pc there affordable and work well but what the other person said the con is that they can catch virus or get hacked but I feel each has its pros and cons
scoundrel [369]3 years ago
4 0

Answer:

I do prefer a PC over a MacBook or other computers because you can customize the hardware and make it bette ring your own way unlike a MacBook where it’s all a secret and you can’t even try to pry it open and if you do the computer would stop working. You can customize your own features to make it better in your own way. A PC is pretty basic to use but the bad thing is that since the systems are most likely not protected, then many hackers/viruses could get in.

You might be interested in
While reviewing system logs, a security analyst notices that a large number of end users are changing their passwords four times
ANTONII [103]
I think the answer is C
6 0
3 years ago
Write the header file Stadium.h for a Stadium class. The Stadium class has the following data members: 1) an array of 1000 Seat
irakobra [83]

Answer:

Check the explanation

Explanation:

File: Seat.h:

typedef struct Seat

{

  int seatNumber;

  string customerFullName;

  double seatCost;

};

File: Stadium.h

class Stadium

{

  //Data Members

  private:

      Seat seats[1000];

      string stadiumName;

      int numOccupiedSeats;

 

  public:

      Stadium(string); //Constructor

      void assignSeat(double, string);

      void unAssignedSeat(int seatNumber);

      int getNumberOfAssignedSeats();

      double getCostOfSeat(double seatNumber);

}

4 0
2 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
Write a new program called Lab7D that will read strings from a text file and turn them into "encrypted code". Create a bool func
Aliun [14]

Answer:

I am writing a C program for the first part and Python program for the second part of the question.    

<h2>1st program:</h2>

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

#include <stdlib.h>  // used here to access exit function

void main() {  //start of the main() function body

char fname[100], character;

// fname is the name of the input file which contains characters and character //variable is used to store characters

FILE *original, *temporary;  

/* two pointers of File type: original is used for the original input file and temporary is used for a temporary file name aux.txt */

printf("Enter the name of the text file to encrypt : ");

//prompts the user to enter the name of the file to be encrypted

scanf("%s",fname); //reads the name of the file from user

original=fopen(fname, "r"); //open the input file in read mode

if(original==NULL)  {

//displays the following message if the file is empty or does not exists

 printf("Cannot open original file");  

 exit(1);  }  //program exits

temporary=fopen("aux.txt", "w");

//creates a temporary file named aux.txt and open that file in write mode

if(temporary==NULL)  { //if temporary file could not be created

 printf("Cannot create a temporary file");

 fclose(original); //closes the original input file

 exit(2);  }  //program exits

while(1)  {

 character=fgetc(original);  

/*pointer original moves through the input file and gets input from original input file one character at a time and store it in character variable */

 if(character==EOF)  

//when all characters are obtained and pointer is at end of file  {

  break;  //the loop breaks }

 else   {  //if EOF is not yet reached

  character=character+100;  

//add 100 to each character to encrypt the characters

  fputc(character, temporary);

 // fputc() writes a single character at a time to aux file } }

fclose(original);  //closes input file

fclose(temporary); //closes aux file

original=fopen(fname, "w");  //opens input file in write mode

if(original==NULL) { //if file does not exist display following message

 printf("Cannot open the original file to write");

 exit(3);  } //program exits

temporary=fopen("aux.txt", "r"); //open aux.txt file in read mode

if(temporary==NULL)  { //if pointer temporary is NULL

 printf(" Cannot open temporary file to read");

 fclose(original);  //closes input file

 exit(4);  } //program exits

while(1)  {

 character=fgetc(temporary);

//obtains every character from aux file pointed by temporary

 if(character==EOF)  //if end of file is reaced {

  break;   } //the loop breaks

 else   { //if end of file is not reached

  fputc(character, original);  

//puts every character to input file }  }

printf(" %s is encrypted \n", fname);  

//displays this message when input file is successfully encrypted

//closes input and aux text files

fclose(original);

fclose(temporary); }  

Explanation:

The program first asks the user to enter the name of the file. Then the program uses two pointers original for input file and temporary for aux text file. It first opens the file whose name is entered by the user and reads that file by getting each single character using fgetc() until the pointer reaches the end of the file. While reading each character of the input file, it encrypts every character using and puts that encrypted content in temporary file names aux.txt using fputc() until the pointer reaches the end of the file. Lastly all the encrypted strings of the are placed in the original input text file.

<h2>Second program:</h2>

# bool function that accepts character as parameter and returns true if its a #vowel and false otherwise

def isVowel(character):  

   if character.lower() in 'aeiou':  #converts uppercase char to lowercase

       return True  #return true if character is a vowel    

   else:

       return False #returns false if input character is not a vowel

The program has a function isVowel() which takes a character as a parameter to check if that character is a vowel. If character is in uppercase  letters, it handles these characters using lower() method to convert the character to lowercase and then returns true if that character is a vowel otherwise returns false. To check the working of the function you can replace True and False with print statement such as:

if character.lower() in 'aeiou':

       print("It is a vowel")

   else:

       print("It is not a vowel")

And after that you can call this function and pass a character to it as:

isVowel('i')

5 0
3 years ago
Candy Kane Cosmetics (CKC) produces Leslie Perfume, which requires chemicals and labor. Two production processes are available:
Vilka [71]

Answer:

  1. Divide the resources into three parts using the corresponding process 1, process 1, and process 2 formats to maximize the use of the resources.
  2. Get the expected revenue by calculating the product of the total perfume in ounce and the price of an ounce of perfume.
  3. Increase the advertisement hours of the product.
  4. subtract the advert fee from the generated revenue to get the actual revenue.
  5. subtract the cost of production from the actual revenue to get the actual profit.

Explanation:

The get maximum profit, all the resources must be exhausted in production. The labor is divided into a ratio of 1:1:2 ( which is 5000, 5000, 1000), while the chemical units are in the ratio of 2:2:3 (10000,10000,15000). This would produce in each individual processes; 15000, 15000 and 25000 oz, which is a total of 55000 oz of perfume.

The expected revenue is $275000. If 1000oz from the 55000oz of perfume is sold without advertisement, model Jenny's awareness of the perfume increases the demand by 200oz per hour, therefore, 24hours would field 4800oz demanded, which would only take 270 hours to distribute all remaining perfumes.

The cost of production would be $130000 for labor and chemical resources plus the advert cost of $27000 ( 270 hours by 100) which is a total cost of $157000. The actual profit is $118000 ( $275000 - $157000).

3 0
2 years ago
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