`{:(0.04x + 0.02y = 5),(0.5(x - 2) - 0.4y = 29):}`

Solve : 0.04x+0.02y= 5 and 0.5( x-2) -0.4 y = 29

0.4 x + 0.3 y = 1.7 , 0.7 x - 0.2 y = 0.8

0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3

Solve for x and y : 0.4 x - 1.5 y = 6.5 , 0.3 x + 0.2 y = 0.9 .

Solve, using cross - multiplication : 0.4x + 1.5y = 6.5 0.3x + 0.2y = 0.9

(0,2) . (0,2) = (0,4)

The solution of 0.2 (2x - 1)-0.5 (3x – 1) = 0.4 is

Determine which of the following can be probability distributions of a random variable X: (i) X: 0 1 2 (ii) X: 0 1 2 P(X): 0.4 0.4 0.2 P(X): 0.6 0.1 0.2 (iii) X: 0 1 2 3 4 P(X): 0.1 0.5 0.2 -0.1 0.3

Solve following pair of equations by using the substitution method. (5) 0.2x + 0.3y = 13 0.4 = 0.5y = 2.3