A bag contains 7 red marbles, 2 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that both marbles drawn will be blue?A bag contains 7 red marbles, 2 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that both marbles drawn will be blue?

1.2%

Step-by-step explanation:

List all the possible pairs and divide 1 by the sum.

Since there were 86 possible pairs,

1/86 is nearly equal to o.o12

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.