1) marbles altogether = 3+4=7
P(black) = 3/7
2) P(first black marble)= 3/7
second black marble - When you removed 1st one only 2 black marbles are left and altogether marbles that are left = 6
P(second black marble) = 2/6=1/3
P(2 black marbles without replacement) means black and black marbles without replacement.
P(2 black marbles without replacement) =3/7*1/3=1/7
3) P(first black marble)= 3/7
When you put that black marble back, you got again <span>3 black marbles and 4 red marbles in a bag. So,
</span> P(second black marble)= 3/7
P(2 black marbles with replacement)=3/7*3/7= 9/49<span>
</span>
2a/5b = 6
(2a-5b)/5b could be written as: (2a/5b) -(5b/5b)
Repace by its equivalent value (6) -(1) = 5
End behavior: f. As x -> 2, f(x) -> ∞; As x -> ∞, f(x) -> -∞
x-intercept: a. (3, 0)
Range: p. (-∞, ∞)
The range is the set of all possible y-values
Asymptote: x = 2
Transformation: l. right 2
with respect to the next parent function:
Domain: g. x > 2
The domain is the set of all possible x-values
Answer:
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Step-by-step explanation:
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