The N on a map would mean North.
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.
In Greek mythology, the Winged Goddess of Victory.
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
#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(); }
The 'K' being silent in knight and not to be confused with the 'K' for king the letter 'N' is generally accepted so as not to confuse the Knight and King in chess notation .
In chemistry, "K" typically represents the element potassium, whose chemical symbol is K. āNā is not specific to any element; it could represent nitrogen or another element starting with the letter N.
The N on a map would mean North.
If you mean knicker, like trousers, k-n-i-c-k-e-r
What does the N represent in VBBN's
The three numbers on a bag of fertilizer represent the percentage by weight of nitrogen, phosphorus, and potassium, respectively. These numbers are known as N-P-K ratios and indicate the concentration of these essential nutrients in the fertilizer.
for (n=1; n<1000; ++n) { for (sum=0, k=1; k<=n/2; ++k) if (n%k==0) sum += k; if (sum==n) printf ("%d\n", n); }
I d o n ' t k n o w ! :D
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.
k = f(n) = 7n
In Greek mythology, the Winged Goddess of Victory.
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }