What is the probability that you roll a dice and get 3, then get an ace from a deck, then (without replacing the other card) draw a black 2? I Really need help!

Explanation:

1/6 is the probability of rolling a '3' since there is one side we want out of 6 total.

The probability of getting an ace is 4/52 because there are 4 aces out of 52 cards total.

If that card isn't put back, then we have 52-1 = 51 cards left. That means the probability of drawing a black 2 is 2/51. The numerator 2 has nothing to do with the card value of '2'. It has to do with the fact we have 2 black cards (2 of spades, 2 of clubs) out of 51 left over.

We multiply the fractions to get the final answer

(1/6)*(4/52)*(2/51)

(1*4*2)/(6*52*51)

8/15912

1/1989

When using a calculator, we can see that,

1/1989 = 0.0005 = 0.05%

There is a very small chance of all three events happening at the same time.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

answer:   y=2x-3

step-by-step explanation: