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Quantity I: n(s) = ¹⁶C₃ = (16× 15× 14)/(3× 2× 1) = 560 n(E) = ⁷C₃ = (7× 6× 5)/(3× 2× 1) = 35 P(E) = 35/560 = 35/560 = 1/16 Quantity II. n(s) = ¹⁶C₃ = (16× 15× 14)/(3× 2× 1) = 560 n(E) = ⁴C₁× ⁵C₁× ⁷C₁ P(E) = 140/560 = 1/4 Hence, Quantity I < Quantity II
I. x2 - 9x - 52 = 0
II. y2 - 16y + 63 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 41x² - 191x + 150 = 0
Equation 2: 43y² - 191y +...
Between what values of x is the expression 19x - 2x2 - 35 positive?
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the...
If x² + 2x + 9 = (x – 2) (x – 3), then the resultant equation is:
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the...
I. x2 - 4x – 21 = 0
II. y2 + 12y + 20 = 0
l). 2p² + 12p + 18 = 0
ll). 3q² + 13q + 12 = 0
I. y/16 = 4/y
II. x3 = (2 ÷ 50) × (2500 ÷ 50) × 42 × (192 ÷ 12)