In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation
Alyce Corkery
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
P1 or parental
P1-e is an expression, not a formula.
If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%
P1: tt F2: tt
#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function
P1" is the highest short-term rating category for Moody's Investor Service. P1 rating are considered to be of high credit quality
P1 Material is the name given to Carbon Manganese Steel in the engineering industry
ed=(q1-q2)/q1/(p1-p2)/p1
#include<stdio.h> #include<conio.h> int main() { char p[10][5],temp[5]; int i,j,pt[10],wt[10],totwt=0,pr[10],temp1,n; float avgwt; clrscr(); printf("enter no of processes:"); scanf("%d",&n); for(i=0;i<n;i++) { printf("enter process%d name:",i+1); scanf("%s",&p[i]); printf("enter process time:"); scanf("%d",&pt[i]); printf("enter priority:"); scanf("%d",&pr[i]); } for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { if(pr[i]>pr[j]) { temp1=pr[i]; pr[i]=pr[j]; pr[j]=temp1; temp1=pt[i]; pt[i]=pt[j]; pt[j]=temp1; strcpy(temp,p[i]); strcpy(p[i],p[j]); strcpy(p[j],temp); } } } wt[0]=0; for(i=1;i<n;i++) { wt[i]=wt[i-1]+et[i-1]; totwt=totwt+wt[i]; } avgwt=(float)totwt/n; printf("p_name\t p_time\t priority\t w_time\n"); for(i=0;i<n;i++) { printf(" %s\t %d\t %d\t %d\n" ,p[i],pt[i],pr[i],wt[i]); } printf("total waiting time=%d\n avg waiting time=%f",tot,avg); getch(); }
Price elasticity demand formula end point formula epd= [q2-q1/q1]/[p2-p1/p1] midpoint formula epd= [q2-q1/(q2+q1)/2] / [p2-p1/(p2+p1)/2]
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.