Nthe f2 generation of a homozygous round (aa) × homozygous wrinkled (aa) cross in peas, three seeds are chosen at random. what is the probability that two seeds are round and the other is wrinkled?

answer: 21$

step-by-step explanation:

$42 / $14 = 3   (ratio 3: 1)

$7 x 3 = $21

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3

answer: the answer is e

step-by-step explanation:

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.