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Lina20 [59]
3 years ago
13

Help me. its due tonight

Computers and Technology
1 answer:
Novosadov [1.4K]3 years ago
5 0
d- enjoy
b- national service !
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At a local burger chain, a customer became extremely angry. Everyone took notice at the person, who was yelling at anyone he cou
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Answer: Donald, the branch manager displayed people skills  

Explanation: People skills are a combination of behavior and behavioral interactions between people. There are many abilities that fall under the category of people skills, for example, personal effectiveness, interactive skills and negotiation skills. While the customer was shouting, Donald patiently heard and tried to calm the customer which reflected that he respected and valued the customer and kept a positive behavior towards the customer.


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Write a function `has_more_zs` to determine which of two strings contains # more instances of the letter "z". It should take as
nekit [7.7K]

Answer:

// Method's name: has_more_zs

// Parameters are text1 and text2 to hold the two phrases to be tested

public static String has_more_zs(String text1, String text2) {

 // Create and initialize z1 to zero

 // Use z1 to count the number of zs in text1

 int z1 = 0;

 // Create and initialize z2 to zero

 // Use z2 to count the number of zs in text2

 int z2 = 0;

 

 //Create a loop to cycle through the characters in text1

 //Increment z1 by one if the current character is a 'z'

 int i = 0;

 while (i < text1.length()) {

  if (text1.charAt(i) == 'z' || text1.charAt(i) == 'Z') {

   z1 += 1;

  }

  i++;

 }

 //Create a loop to cycle through the characters in text2

 //Increment z2 by one if the current character is a 'z'

 

 i = 0; //Re-initialize i to zero

 

 while (i < text2.length()) {

  if (text2.charAt(i) == 'z' || text2.charAt(i) == 'Z') {

   z2 += 1;

  }

  i++;

 }

 

 

 //Using the values of z1 and z2, return the necessary statements

 if (z1 > z2) {

  return "The phrase '" + text1 + "'" + " has more occurences of z than the phrase " + "'" + text2 + "'";

 }

 else if (z1 < z1) {

  return "The phrase '" + text2 + "'" + " has more occurences of z than the phrase " + "'" + text1 + "'";

 }

 else if (z1 == z2) {

  return "The strings have the same number of z";

 }

 else {

  return "Neither string contains the the letter z";

 }

}

Explanation:

Explanation to answer has been given in the code as comments. Please refer to the comments for the details of the code.

The source code file has been attached to this response and saved as "NumberOfZs.java"

Hope this helps!

Download java
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In python how do you write for prime factors
olasank [31]

Answer:

The while loop is used and the factors of the integer are computed by using the modulus operator and checking if the remainder of the number divided by i is 0. 3. Then the factors of the integer are then again checked if the factor is prime or not.

Explanation:

google

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You are !o code and execute a C program to input narnes ard addresses t]rat are in alphabctic order and output the names and add
alexgriva [62]

Answer:

#define MAX 50

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

struct address

{

char name[MAX];

char street[MAX];

char city[MAX];

char zip[10];

};

void in(struct address *, int *);

void sort(struct address *, int);

void out(struct address *, int);

int main(){

struct address *ADDRESS = (struct address *)malloc(50*sizeof(struct address));

int n=0;

in(ADDRESS, &n);

sort(ADDRESS, n);

out(ADDRESS, n);

}

void in(struct address *ADDRESS, int *num)

{

FILE *fp;

char fname[MAX];

printf("Enter input file name: ");

scanf("%s", fname);

fp = fopen(fname, "r");

if (fp == NULL)

{

printf("Unable to open file");

exit(1);

}

int a = 0;

while (!feof(fp))

{

//ADDRESS = (struct address *)realloc(ADDRESS, (a + 1) * sizeof(struct address));

if(a>=50)

break;

fscanf(fp, "\n%[^\n]s", (ADDRESS+a)->name);

fscanf(fp, "\n%[^\n]s", (ADDRESS+a)->street);

fscanf(fp, "\n%[^\n]s", (ADDRESS+a)->city);

fscanf(fp, "\n%[^\n]s", (ADDRESS+a)->zip);

a++;

}

printf("Successfully %d read from file\n", a);

*num = a;

}

void sort(struct address *ADDRESS, int num){

int i, j;

for(i=0; i<num; i++){

for(j=i+1; j<num;j++){

if(strcmp((ADDRESS+i)->zip, (ADDRESS+j)->zip)>0){

struct address t = *(ADDRESS+i);

*(ADDRESS+i) = *(ADDRESS+j);

*(ADDRESS+j) = t;

}

}

}

}

void out(struct address *ADDRESS, int num)

{

FILE *fp;

char fname[MAX];

printf("Enter output file name: ");

scanf("%s", fname);

fp = fopen(fname, "w");

if (fp == NULL)

{

printf("Unable to open file");

exit(1);

}

int a;

for(a=0; a<num; a++)

{

fprintf(fp, "%s\n", (ADDRESS+a)->name);

fprintf(fp, "%s\n", (ADDRESS+a)->street);

fprintf(fp, "%s\n", (ADDRESS+a)->city);

fprintf(fp, "%s\n", (ADDRESS+a)->zip);  

}

printf("Successfully %d Write to file\n", a);

fclose(fp);

}

Explanation:

4 0
3 years ago
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