The number of errors in a textbook follow a poisson distribution with a mean of 0.02 errors per page. what is the probability that there are 3 or less errors in 100 pages?

0.8571

Step-by-step explanation:

Poisson distribution formula:

 , where x=0,1,2,3... is a poisson variable , = Mean of the distribution .

Let x be the number of errors in a textbook that follow a Poisson distribution with a mean of 0.02 errors per page.

For 100 pages ,

Now , the probability that there are 3 or less errors in 100 pages will be :-

Hence, the probability that there are 3 or less errors in 100 pages = 0.8571

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.