Q: What is 10 n on a p?

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The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.

Multiplying by a negative changes the sign of the original number. P x N = N N x N = P Multiplying by a positive keeps the sign of the original number. P x P = P N x P = N

Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.

p+n=x

E= P x r x (1 + r)n / ((1+r)n -1) Here p=principal amount r = interesr rate per month(ex: if interest rate per annum is 10% then 10/(12*100)) n= tenure in months

Related questions

#include<stdio.h> void main() { int *p,n,s=0; printf("Enter Number :"); scanf("%d",&n); for(p=&n;*p>0;p++) { s=s*10+*p%10; *p=*p/10; } printf("Reverse Number=%d",s); }

Try 151,600! Permutations & Combinations. P(n,r)=n!(n−r)! not P(n,r)=n!/(n!-r!)r! ?

#include<stdio.h> #include<conio.h> void main() { clrscr(); int r=0,d,m,n; printf("Enter a value="); scanf("%d",&n); m=n; do { d=m%10; m=m/10; r=(r*10)+d; } while(m!=0); printf("%d is the reverse",r); getch(); }

I'll use these symbols for each coin: P = Penny; D = Dime; N = Nickel 12 P 7 P & 1 N (7 + 5) 2 P & 2 N (2 + 10) 1 D & 2 P (10 + 2)

P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!

N - p% = N - p% of N = N*(1 - p%) = N*(1 - p/100) or N*(100 - p)/100

BJT is nothing but the addition of two PN junction diodes. There are two types of BJT= P-N-P or N-P-N P-N N-P + or + N-P = P-N-P P-N =N-P-N SCR is a thyristor which is made adding two BJTs. Of course they are made of sillicon. Exempli gratia: P-N-P + + N-P-N = P-N-P-N comparison between scr bjt and mosfet Check the related link for further information.

The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.

Positive. p*p=p p*n=n n*n=p

#include<stdio.h> #include<conio.h> main() { int a[10][10],b[10][10],c[10][10],m,n,i,j,p,q,op; printf("enter the order of matrix a:n"); scanf("%d",&m,&n); printf("enter the %d elements of a\n",m*); for(i=0;i<m;i++) for(j=0;j<n;j++) scanf("%d",&a[i][j]); printf("enter the order of matrix b:n"); scanf("%d",&p,&q); printf("enter the %d elements of b\n",p*q); for(i=0;i<p;i++) for(j=0;j<q;j++) scanf("%d",&b[i][j]); printf("enter the option\n"); scanf("%d",&option); switch(op) { case '+' : if(m==p&&n==q) printf("the resultant matrix c is:\n"); for(i=0;i<m;i++) for(j=0;j<n;j++) c[i][j]=a[i][j]+b[i][j]; printf("%d",c[i][j]); printf("\n"); break; case '/' : if(n==p) { for(i=0;i<m;;i++); {for(j=0;j<q;j++) {printf("%d",c[i][j]); } } c[i][j]=0; for(p=0;p<n;p++) c[i][j]=c[i][j]+a[i][p]*b[p][j]: } printf("resultant matrix is:\n"); for(i=0;i<m;;i++); {for(j=0;j<q;j++) {printf("%d\t",c[i][j]); } } printf("\n"); getch(); }

normal variable stores a value of the given datatype where as the pointer variable stores the address of a variable. for example int n=10; int *p; p=&n; here p is a pointer variable and n is a normal variable.p stores the address of n where as n stores an integer value. *p prints the value of n,p prints the address of n.

A negative times a negative is a positive. A simple way to remember this is... n * p = n p* n = n p * p = p n * n = p There will always be two negatives, and two positives.