Answer:
Débito o debe Crédito o haber
Préstamo de la cooperativa Efectivo
Gasto de arriendo Cuentas por cobrar
Cuentas por pagar Documentos por cobrar
Gasto de publicidad Capital social
Hipoteca por pagar Ventas
Intereses pagados Bancos (en el caso de cobros a través de bancos)
Bancos (en el caso que deba productos bancarios)
Explanation:
I should be skeptical of the credibility of sources that I have during the gathering. Be cautious and make sure that data comes of from trusted specialized sources. Trusted sources can be easily identified for their popularity on certain fields. Determine the purpose of the site and the data it contains.
Answer:
The code to this question can be given as:
Code:
int i,j,count_previous=0,count_next=0; //define variable
for (j=0; j<n; j++) //loop for array
{
if (x[0]==x[j]) //check condition
{
count_previous++; //increment value by 1.
}
}
for (i=0; i<n; i++) //loop
{
for (j=0; j<n; j++)
{
if (x[i]==x[j]) //check condition
{
count_next++; //increment value by 1.
}
}
if (count_previous>count_next) //check condition
{
mode=x[i-1];
}
else
{
mode=x[i];
count_previous=count_next; //change value.
count_next= 0 ; //assign value.
}
}
Explanation:
In the question it is define that x is array of the string elements that is already define in the question so the code for perform operation in the array is given above. In the code firstly we define the variable that is i, j, count_previous, count_next. The variable i,j is used in the loop and variable count_previous and count_next we assign value 0 that is used for check the values of array. Then we define the for loop in this loop we use conditional statement in the if block we check that array of zero element is equal to array of j value then the count_previous is increase by 1. Then we use nested loop It is also known as loop in a loop. In this first loop is used for array and the second loop is used for check array element.In this we use the condition that if array x of i value is equal to array x of j then count_next will increment by 1.Then we use another condition that is if count_previous is greater then count_next then mode of x is decrement by 1. else mode equal to array and count_previous holds the value of count_next and count_next is equal to 0.
Answer:
To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.
Similarly check for 400th year and multiple 0f 4. The following C program describes the function.
#include<stdio.h>
#include<stdbool.h>
bool is_leap_year(int year);
void main()
{
int y;
bool b;
printf("Enter the year in yyyy format: e.g. 1999 \n");
scanf("%d", &y); // taking the input year in yyyy format.
b= is_leap_year(y); //calling the function and returning the output to b
if(b==true)
{
printf("Thae given year is a leap year \n");
}
else
{
printf("The given year is not a leap year \n");
}
}
bool is_leap_year(int year)
{
if(year%100==0) //every 100th year
{
if(year%400==0) //every 400th year
{
return true;
}
else
{
return false;
}
}
if(year%4==0) //is a multiple of 4
{
return true;
}
else
{
return false;
}
}
Explanation:
Output is given as image
