Question
A vessel contains 5 red, βxβ green and 3 blue
ribbons. If two ribbons are drawn from the vessel randomly, then the probability of getting a red and a green ribbon is 3/11. Find the value of x.Solution
Total number of ribbons = (x + 8) According to question; (xC1 Γ 5C1)/(x + 8)C2 = 3/11 2 Γ 5x/(x + 8)(x + 7) = 3/11 110x = 3 Γ (x2 + 15x + 56) 110x = 3x2 + 45x + 168 3x2 β 65x + 168 = 0 3x2 β 9x β 56x + 168 = 0 3x(x β 3) β 56(x β 3) = 0 (3x β 56)(x β 3) = 0 So, x = 56/3 or x = 3 Since number of ribbons canβt be in fraction Therefore, x = 3
(γ(0.5)γ^(1/3)Β Γ γ(1/125)γ^(1/4)Β Γ γ25γ^(1/6)Β Γγ(6.25)γ^(2/3))/(γ(2.5)γ^(2/3)Β Γ 5^(-1/2)Β Γ γ(1/5)γ^(-2)Γγ3125οΏ½...
?2 - (40% of 240) = 25 X 5
β157464 =?
32 + 26 Γ (484 Γ· 44) + 450 Γ· 9 = ?Β
(1748 ÷ 8) + 76.8 × 35 =(? × 4) + (42 × 35.5)
52% of 36% of 810 = 72% of 18% of ?Β
Find the simplified of the following expression:
Solve:
5(x - 2) = 3(x + 4) + 4
18 * 9 + 25% of 120 + 50% of 200 = ?
756 + 432 β 361 + ? = 990
