0
What else can I help you with?
What is the formula for euqated monthly installment of loan amount?
Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12 where l = loan amt r = rate of interest n = no of terms
What is the formula to calculate mortgage payment?
The answer: you probably don't want to know. Here it is: P = L[c(1 + c)n]/[(1 + c)n- 1] where P equals payment, L equals loan amount, n equals number of months in the loan term and c equals monthly interest rate. If you are just wanting to calculate what a monthly mortgage payment would be, there are several mortgage calculators available online. Here's a link to a good one: http://www.bankrate.com/calculators/mortgages/mortgage-calculator.aspx
What is a C E L N loan?
Continuing Education Loan (private)
Formula for recurring deposit?
R.D.=P*n+P(n+1)/2400
What is the formula for PVIFA?
PVIFA = (1 - 1 / (1 + r)n) / r where n is the number of payment periods; r is the nominal interest rate for one period
Why is it said that poisson distribution is a limiting case of binomial distribution?
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
What does the formula n(1-R)L mean?
n(1-R)L is an expression: it is not a formula.
Values of n l and ml of 2p orbital?
n : 2 l : 1 ml : -1, 0, or 1
How many possible values for l and ml are there when n equals 4?
(N-1)=(4-1)= N=3 l=0,1,2,3
Prove that the sequence n does not converge?
If the sequence (n) converges to a limit L then, by definition, for any eps>0 there exists a number N such |n-L|N. However if eps=0.5 then whatever value of N we chose we find that whenever n>max{N,L}+1, |n-L|=n-L>1>eps. Proving the first statement false by contradiction.
What is the formula for euqated monthly installment of loan amount?
Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12 where l = loan amt r = rate of interest n = no of terms
Longest common subsequence problem program in c?
#include<stdio.h> #include<string.h> int max(int a,int b) { return a>b?a:b; }//end max() int main() { char a[]="xyxxzxyzxy"; char b[]="zxzyyzxxyxxz"; int n = strlen(a); int m = strlen(b); int i,j; for(i=n;i>=1;i--) a[i] = a[i-1]; for(i=m;i>=1;i--) b[i] = b[i-1]; int l[n+1][m+1]; printf("\n\t"); for(i=0;i<=n;i++) { for(j=0;j<=m;j++) { if(i==0 j==0) l[i][j]=0; else if(a[i] == b[j] ) l[i][j] = l[i-1][j-1] + 1; else l[i][j] = max(l[i][j-1],l[i-1][j]); printf("%d |",l[i][j]); } printf("\n\t"); } printf("Length of Longest Common Subsequence = %d\n",l[n][m]); return 0; }
How many possible combinations are there for the values of l and ml when n2?
For the principal quantum number ( n = 2 ), the possible values of the azimuthal quantum number ( l ) are 0 and 1 (since ( l ) can take on values from 0 to ( n-1 )). For each value of ( l ), the magnetic quantum number ( m_l ) can take values from (-l) to (+l). Therefore, for ( l = 0 ), ( m_l = 0 ) (1 combination), and for ( l = 1), ( m_l ) can be (-1, 0, +1) (3 combinations). In total, there are ( 1 + 3 = 4 ) possible combinations of ( l ) and ( m_l ) for ( n = 2 ).
What value of secondary quantum number l?
The secondary quantum number, l, represents the shape of an orbital and can have values ranging from 0 to n-1, where n is the principal quantum number. Therefore, l can have values from 0 to (n-1).
How many values of l are possible when n equals 5?
For an electron with n=5, the possible values of l range from 0 to 4 (l=0, 1, 2, 3, 4). The value of l depends on the principal quantum number (n) according to the rule that l can be any integer value from 0 to n-1.
What is 1I the L N?
1 is the lonliest number
What are the principal quantum number and the azimuthal quantum number in an atom of Potassium?
Potassium has an electronic configuration of 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 The azimuthal quantum number of 0corresonds to an s subshell, 1 p subshell You can see that you have n=1, l=0 n=2, l=0; n=2, l=1 n=3, l=0; n=3, l=1, n=4, l=0
