11 + k
That depends on the value assigned to the variable "k". If you don't know the value of "k", you express the sum as 11 + k; in this case, you can't simplify it any more.
k12
(9+k)-17=x
100/(k+m)Its actual value depends on the values of 'k' and 'm' .
17
The sum of the even numbers up to 2k, where k is an integer, is k(k + 1) = k2 + k
The flowchart to read 10 positive integers K>10 Start A N K=1 Sum = 0 Sum = Sum + K2 B Is Y Print K > 100? sum K=k+1 End B A
Print "Type the upper limit (n) ?" Input n K = -1 WHILE K < = n K = K + 2 Sum = Sum + K WEND Print "The sum of all odd numbers up to "; n; "is "; Sum
11 + k
k + 7
enter the number whose digits are to be added num is the given value num=0! k=num%10 sum=sum=k k=num/10 num=k print the sum of the digits
it would have a part in it like this: for (i=0; i<n; ++i) { . for (j=0; j<l; ++j) { . . sum= 0; . . for (k=0; k<m; ++k) { . . . sum += a[i][k] * b[k][j]; . . } . . c[i][j] = sum; . } }
k + 7
2(k+4)
If X has the Poisson distribution with mean l then Pr(X = k) = e-llk/k! Mean of Poisson = Sum over all k of [k*P(X = k)] which happens to be l. = Sum over all k of [k*e-llk/k!] = Sum over all k of [e-llk/(k-1)!] = Sum over all j of [le-llj/j!] where j has been substituted for k-1 = l*Sum over all j of [e-llj/j!] But the quantity being summed is simply the pdf of the Poisson distribution and so its sum over all possible values is 1 So Mean = l And then Variance of Poisson = Sum over all k of [k2*P(X = k)] - l2. = Sum over all k of [k2*e-llk/k!] - l2 Then, since k2 = k*(k-1) + k Variance = Sum over all k of [k*(k-1)e-llk/k!] +Sum over all k of [k*e-llk/k!] - l2 = Sum over all j of [l2e-llj/j!] where j has been substituted for k-2 + Sum over all i of [le-lli/i!] where i has been substituted for k-1 - l2 = l2*Sum over all j of [e-llj/j!] + l*Sum over all i of [e-lli/i!] - l2 And since the sums are equal to 1, Variance = l2 + l - l2 = l Apologies: I did the answer using the symbol for lambda but this browser changed them all back to l. I cannot change them all t o something else but I hope it is clear. At least they are distinguishable from 1!
k^2 + k = k^2 + k k^2 x k = k^3