NPK refers to the three essential nutrients in fertilizers: nitrogen (N), phosphorus (P), and potassium (K). Nitrogen promotes healthy leaf and stem growth, phosphorus supports root development and flowering, while potassium enhances overall plant health, disease resistance, and fruit quality. Together, these nutrients help improve crop yields, enhance soil fertility, and promote balanced plant growth, making NPK fertilizers critical for agricultural productivity.
P. K. N. Burbidge has written: 'Tort'
N. K. P. Salve was born on 1921-03-18.
N. K. P. Salve died on 2012-04-01.
P = 1 For K = 1 to M . P = P * N Next K PRINT "N raised to the power of M is "; P
You need a formula for this. If the probability (in one toss) of getting head is "p", then the probability of getting exactly k heads out of n tosses is: (n,k) p^k (1-p)^(n-k) where (n,k) denotes the number of combinations of k elements among n. You should also know that (n,k) = n! / (( n-k)! k! ) so here, with n=8, k=6, and p=.5 you have (n,k) = 8*7 / 2 = 28 and your probability is : 28 * 1/2^6 * 1/2^2 = 28 / 256 = 7 / 64
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
p-i-n-k
That is the N P K numbers. N=Nitrogen P=Phosphorous K=Potassium. the numbers are always in that order and tell you the percentage.
P-r-a-n-k.
/* Discrete Fourier Transform and Power Spectrum Calculates Power Spectrum from a Time Series Copyright 1985 Nicholas B. Tufillaro */ #include #include #define PI (3.1415926536) #define SIZE 512 double ts[SIZE], A[SIZE], B[SIZE], P[SIZE]; main() { int i, k, p, N, L; double avg, y, sum, psmax; /* read in and scale data points */ i = 0; while(scanf("%lf", &y) != EOF) { ts[i] = y/1000.0; i += 1; } /* get rid of last point and make sure # of data points is even */ if((i%2) == 0) i -= 2; else i -= 1; L = i; N = L/2; /* subtract out dc component from time series */ for(i = 0, avg = 0; i < L; ++i) { avg += ts[i]; } avg = avg/L; /* now subtract out the mean value from the time series */ for(i = 0; i < L; ++i) { ts[i] = ts[i] - avg; } /* o.k. guys, ready to do Fourier transform */ /* first do cosine series */ for(k = 0; k <= N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*cos(PI*k*p/N); } A[k] = sum/N; } /* now do sine series */ for(k = 0; k < N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*sin(PI*k*p/N); } B[k] = sum/N; } /* lastly, calculate the power spectrum */ for(i = 0; i <= N; ++i) { P[i] = sqrt(A[i]*A[i]+B[i]*B[i]); } /* find the maximum of the power spectrum to normalize */ for(i = 0, psmax = 0; i <= N; ++i) { if(P[i] > psmax) psmax = P[i]; } for(i = 0; i <= N; ++i) { P[i] = P[i]/psmax; } /* o.k., print out the results: k, P(k) */ for(k = 0; k <= N; ++k) { printf("%d %g\n", k, P[k]); } }
/*PROGRAM TO IMPLEMENT GAUSS-jordan method.#include#define MX 20main(){float a[MX] [MX+1],m,p;int i,j,k,n;puts("\n how many equations?:");scanf("%d",&n);for(i=0;i
Mobil-1 p/n M1-103 Wix p/n 51394 Amzoil p/n EA15K09 K & N p/n KN-128 Fram p/n PH4967