Abox contains only black and white chips. there are 8 black chips and 6 white chips. how many white chips must be removed so that the possibility of randomly picking a black chip becomes 2/3 ? a) 1 b) 2 c) 3 d) 4 me pl whoever answers this right gets the brainliest

There are 8 black chips and 6 white chips, 14 chips in total. In this situation the probability to pick a black chip is 8/14=4/7.

When one white chip was removed, then only 5 white chips left and the total number of chips became 13. In this situation the probability to pick a black chip is 8/14=4/7.

When second white chip was removed, then ther left 4 white chips and the total number of chips became 12. Now the probability to pick a black chip is 8/12=2/3 as needed.

Must be removed 2 white chips. Correct choice is B.

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation:

p - 6

step-by-step explanation

so you know that it cost $6 to see a movie you subtract p - 6 to get your answer

hope this !