Answer:
The probability of selecting randomly 6 packets such that the first 4 contain cocaine and 2 not is 0.3522.
Step-by-step explanation:
Probabilities of selecting 4 packets with the ilegal substance:
= 
Combinassions possible= 1365
Probabilities of selecting 2 packets with white powder:
= 
Combinations possible= 10
Probabilities of selecting 6 packets from the totality of them:
= 
Combinations possible= 38760
The probability of picking 4 with the substance and 2 with only white powder is:
= 0.3522
I hope this answer helps you.
It is rational because it can be represented by p/q form where p and q are whole number
Answer:
A-33(PAY) OR 33(P)
B-33 x pay or 33 x p
C-hours(33)or H(33)
D-pay(33) or p(33)
its D is correct hope this helps
Well the measure of angle 1 in problem 1 is 62, in problem 2 is 90, problem 3 is 45, and problem 4 is 60
Distribute the .25 to get .25x + 1 - 3 = 28
Then simplify to get .25x - 2 = 28
Move the two to the right to get .25x = 30
Then divide to leave x: 30 / .25 = 120
X = 120