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mestny [16]
3 years ago
14

How many levels of full body protective clothing are there? A. Four B. Seven C. Six D. Two

Computers and Technology
2 answers:
DaniilM [7]3 years ago
6 0
<h2>Answer</h2>

Option A - Four

<u>Explanation</u>

Full body protective clothing or PPE, provides protection from serious injuries or diseases. It is very important for full body to have it as it is common within human being. In addition to this, the research has shown that protection is usually divided into various categories. It is divided into four categories based on how much the protection is needed.

Verdich [7]3 years ago
4 0
The answer is A. Four
You might be interested in
Write a function chat accepts a pointer to a C-string as its argument. The function should count the number of vowels appearing
Dovator [93]

Answer:

Function contains into the code asked for:

  • Receives a string as an argument.  
  • count the number of vowels in the string and return that number.
  • receives a string as an argument.
  • count the number of consonants in the string and return that number.  

Explanation:

//Let's call libraries

#include <stdio.h>  

#include <conio.h>

//Start main

int main()

{

//*Declare functions for counting number of vowels and consonants

int vowels=0;      

int consonants=0;    

int all=0;    

//Output need information

//Ask user to input a string

printf("what do you want to count?\n");    

printf("1-the vowels.\n");      

printf("2-the consonants.\n");    

printf("3-all of the letters on a phrase.\n");      

printf("\n");      

printf("-> ");      

int a;    

a = getchar();

   switch (a)

   {          

//Ask user what function they want to run

case '1':

//Call the function needed to count the vowels  

            printf("Introduce a phrase: ");                  

char textoa[1001];                

scanf("%s", &textoa);

char *p;                  

p = textoa;                

while (*p != '\0')

               {

                     if (*p == 'a' || *p == 'e' || *p == 'i' || *p == 'o' || *p == 'u' || *p == 'A' || *p == 'E' || *p == 'I' || *p == 'O' || *p == 'U') vocales ++;

                     p++;

               }

               printf("there are %d vowels.", vowels);                  

break;            

case '2':

//Call the function needed to count the consonants

printf("Introduce a phrase: ");                

char textob[1001];                

scanf("%s", &textob);

p = textob;                

while (*p != '\0')

               {

                     if (*p == 'b' || *p == 'B' || *p == 'c' || *p == 'C' || *p == 'd' || *p == 'D' || *p == 'f' || *p == 'F' || *p == 'g' || *p == 'G' || *p == 'h' || *p == 'H' || *p == 'j' || *p == 'J' || *p == 'k' || *p == 'K' || *p == 'l' || *p == 'L' || *p == 'm' || *p == 'M' || *p == 'n' || *p == 'N' || *p == 'p' || *p == 'P' || *p == 'q' || *p == 'Q' || *p == 'r' || *p == 'R' || *p == 's' || *p == 'S' || *p == 't' || *p == 'T' || *p == 'v' || *p == 'V' || *p == 'w' || *p == 'W' || *p == 'x' || *p == 'X' || *p == 'y' || *p == 'Y' || *p == 'z' ||  

*p == 'Z') consonants ++;

                     p++;

               }

               printf("there are %d consonants.", consonants);                  

break;            

case '3':

printf("Introduce a phrase: ");                

char textoc[1001];                

scanf("%s",&textoc);

p = textoc;                

while (*p != '\0')

               {

                     if (*p == 'a' || *p == 'e' || *p == 'i' || *p == 'o' || *p == 'u' || *p == 'A' || *p == 'E' || *p == 'I' || *p == 'O' || *p == 'U' || *p == 'b' || *p == 'B' || *p == 'c' || *p == 'C' || *p == 'd' || *p == 'D' || *p == 'f' || *p == 'F' || *p == 'g' || *p == 'G' || *p == 'h' || *p == 'H' || *p == 'j' || *p == 'J' || *p == 'k' || *p == 'K' || *p == 'l' || *p == 'L' || *p == 'm' || *p == 'M' || *p == 'n' || *p == 'N' || *p == 'p' || *p == 'P' || *p == 'q' || *p == 'Q' || *p == 'r' || *p == 'R' || *p == 's' || *p == 'S' || *p == 't' || *p == 'T' || *p == 'v' || *p == 'V' || *p == 'w' || *p == 'W' || *p == 'x' || *p == 'X' || *p == 'y' || *p == 'Y' || *p == 'z' || *p == 'Z') all ++;

                     p++;

               }

               printf("there are %d letters on the phrase.", all);                 break;

   }     getch();

}

8 0
3 years ago
How do you get free Wifi on your phone without paying
Monica [59]
The library offers free wifi. I don’t think you can get free Wifi without paying(unless you go somewhere that offers free wifi). I don’t recommend you hacking into people’s Wifi(you can go to jail).
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Which option identifies the programming paradigm selected in thr following scenario? A student is writing a science fiction stor
NISA [10]

Answer:

java

Explanation:

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All of the following items could be too expensive for someone with only a high school degree to afford except
Papessa [141]
This sounds like a multiple choice answer can I have 4 choices to pick?
7 0
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anyanavicka [17]
Ubuntu, Linux, and Mint
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