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What does n on a map normally represent?
The N on a map would mean North.
What are the degrees of freedom n-k in statistics?
The degrees of freedom is a count of the number of variables which are free to vary. This number can be fewer than the number of variables because of constraints imposed by additional information. Suppose, for example, you have n variables and you also know their mean. Knowing the mean is effectively the same as knowing their sum. So (n-1) of the variables can take any value but, after that, the nth variable must be such that all n of them add to the know sum. Similarly, if you have n variables and are given k means (for k classes), then only n-k of the variables are free to take any value. There are k constraints imposed by the k means.
What does N 1 K E Mean?
In Greek mythology, the Winged Goddess of Victory.
Prove that 2 power n lessthan equal to n factorial lessthan equal to n power n?
Assume 2^k < k! for all n > k here n > 2, then 2^n = 2^(n - 1)*2 < (n-1)! * n = n! Done. Connie and John
Gaussian elimination in c?
#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(); }
