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the question is = The S R of -1 is an I N = so

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the square root of -1 is an imaginary number

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How do you find the sum of an infinite geometric series?

An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)


With this radical n equals S r exponent-2 solve for r?

sqrt(n) = S*r2sqrt(n)/S = r2 sqrt(sqrt(n)/S) = ror4th root of n/sqrt(S) = r.If your equation was sqrt(n) = S*r-2sqrt(n) = S*(1/r2)sqrt(n)/S = 1/r2S/sqrt(n) = r2 sqrt(S/sqrt[n]) = rorsqrt(S)/4th root of n = r


Prove that nCr plus nCr minus 1 equals n plus 1Cr?

nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr


Implimentation of pascle triangle in C?

#include#includevoid main(){int n,a=1,s=1,r;printf("\n enter number of lines ");cin>>n;for(;a=1;b--){printf(" ");}r=pow(s,2);printf("\n");a++;}getch();}// \m/


The S R of -1 is an 1 N?

the square root of -1 is an imaginary number


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.


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Sirens is spelled S-I-R-E-N-S.


The s r of - 1 is an you n?

Square root of -1 is mathematically represented by the IMAGINARY NUMBER i.


What is the difference between an arithmetic series and a geometric series?

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The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n. and where nCr = n!/[r!*(n-r)!]


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#include&lt;iostream&gt; using namespace std; int revDigit(int n){ int r,s=0; while(n&gt;0){ r=n%10; n=n/10; s=(s*10)+r; } return s; } main(){ long n,r,r1,s=0,s1=10; cout&lt;&lt;"Enter a number."&lt;&lt;endl; cin&gt;&gt;n; while(n&gt;0){ r=n%2; n=n/2; s=(s*10)+r; } //since we read the binary equivalent in a bottom up manner, we need to reverse it. s1=revDigit(s); cout&lt;&lt;"Binary equivalent:\t"&lt;&lt;s1&lt;&lt;endl; system("pause");//this line is required if you are using dev c++ }