Answer:
BUTTHOLE IS A BUTT THAT FARTS POOP....YA
The probability that a randomly chosen person from this group has type B is <u>3/25</u>
The probability that a randomly chosen person from this group has type AB is <u>1/25</u>
The probability that a randomly chosen person from this group has type B or type AB blood is <u>4/25</u>
Answer:
Step-by-step explanation:
−189=4x−3(−4x+15)
−189=16x−45
16x−45=−189
16x−45+45=−189+45
16x=−144
16x
/16
=
−144
/16
x=−9
Good luck honey !!
Step #1 for both: figure out which interval your x-value fits into.
For f(-2), x=-2 and -2 fits with x ≤ -2, the top interval.
For f(3), x=3 and 3 fits into -2 < x ≤ 3, the middle interval.
Step #2 for both, plug in your x-value to the piece of the function that fits with that interval.
For f(-2), we know x≤-2, so we use 2x+8 to evaluate x=-2.
For f(3), we know -2
f(-2) = 2(-2)+8 = -4+8 = 4
f(3) = (3)^2 -3 = 9-3 = 6
Let’s say, hypothetically speaking, you chose the second marble without replacing the first marble so, events are hypothetically dependent. Events are dependent if the occurrence of one hypothetical event hypothetically does affect the likelihood that the other events occur. The probably of two or more dependent events A and B is the probability of A times the probability of B after A hypothetically occurs
P(A and B) = P(A) x P(B after A)
Choose the first marble
The total number of hypothetical marbles are, hypothetically speaking, 4 on a hypothetical basis, and there is one red marble.
P(red)=1/4
Choose the second marble
Without hypothetically replacing the hypothetical first marble, you choose the hypothetical marble, hypothetically speaking. So, the total hypothetical number of marbles are, hypothetically, 3, and there is, hypothetically, one green marble.
P(green) = 1/3
The probability of choosing red and then, hypothetically, green is:
P(red and green) = P(red) x P(green)
=1/4 x 1/3
= 1/12
P(red and green) is hypothetically equal to 1/12 on a hypothetical account.
Final hypothetical answer: 1/12
