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.
Add all the numbers up then divide that by 20 that’s your answer. I will not give you the answer because you also need to learn sorry :)
Answer:
15(x - 5)(x - 3)²
Step-by-step explanation:
factor both denominators and see what you need to do to make them equal...
(x² - y²)/[(x - 5)(x - 3)] y²/[15(x - 3)(x - 3)]
the first denominator has (x - 5)(x - 3)
the second denominator has 15(x - 3)(x - 3),
the least common denominator between them would be
15(x - 5)(x -3)(x - 3) or 15(x - 5)(x - 3)²
Answer:
13
Step-by-step explanation:
-2(-2) - (-8) + 1
4 + 8 + 1 = 13
Answer:
285
Step-by-step explanation:
use algebra
third is x
second is 4x
first is 4x-3
x+4x+4x-3
9x-3 = x+x+1+x+2
9x-3=3x+3
12x=6, divide x=2
third is 2, sec is 8, first is 5
