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