Abag contains 3 red, 4 blue, 3 green, and 5 yellow marbles. what is the probability of
picking two green marbles in a row, without replacing the first marble?

1/35 ≈2,85%

Step-by-step explanation:

Total possible outcomes for the first pick: 3+4+3+5 = 15

Probability of picking a green marble for the first pick : 3/15 = 1/5

Total possible outcomes for the second pick: 15-1(the first pick) = 14

Probability of picking a green marble for the second pick : 3-1(first pick)/14 = 2/14 = 1/7

Probability of picking green marble 2 times in a row = 1/5 * 1/7 = 1/35 ≈2,85%

Hope this help you :3

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.