Solve: `2 (3k + 7) + 3 ( 2k -5) = 23`

Solve : 2(k + 3) + 3 (k-4) = 24

lim_ (n rarr oo) sum_ (k = 1) ^ (n) (lambda k ^ (4) + 2k ^ (3) + k ^ (2) + k + 1) / (3n ^ (5) + n ^ (2) + n + 5k) = (1) / (3)

Solve for k : 1.5 k -3.7 = 0.8

(17-2k)/(k-5)=-3

If the equation (k^(2)-3k +2) x^(2) + ( k^(2) -5k + 4)x + ( k^(2) -6k + 5) =0 is an identity then the value of k is

5x + 2y = 2k , 2 (k + 1 ) x + ky = (3k + 4)

A random, variable X has the following probability distribution: X : 0 1 2 3 4 5 6 7 P(X):0 k 2k 2k 3k k^2 2k^2 7k^2+k Find each of the following: i. k ii. P(X<6) iii. P(Xgeq6) iv. P(0< X < 5)

A random variable X has the following probability distribution: I 0 1 2 3 4 5 6 7 P\ (X) 0 K 2K 2K 3K K^2 2K^2\ 7K^2+K Determine: (i) K (ii) P(X 6) (iv) P (0 < X < 3)

For what value of k, will the following pair of equations have infinitely many solutions: 2x+ 3y=7 and (k+2)x-3 (1-k) y= 5k +1