Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is 1 star per 16 cubic light years. (a) What is the probability of 3 or more stars in 16 cubic light years?(b) How many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.92?

a) 0.264

b) 48 cubit light-years

Step-by-step explanation:

For calculating a)  What is the probability of 3 or more stars in 16 cubic light years?

a)

λ= 1 star/16 cubic light-years

t= measure t in units of 16 cubic light years.

E(Y) = λ t = (1/16)(16) = 1 star

P(X>=2) = 1 - P(X<2)

= 1 - [e^-1 + (e^-1)(1)/1!]

= 0.264

b)

P(X≥1) = 1 - P(X=0)

= 1 - e^-μ

0.95 = 1 - e^-μ

e^-μ = 1 - 0.95

e^-μ = 0.05

ln(e^-μ) = ln(0.05)

-μ = -3

μ = 3

Therefore  3 x 16 = 48 cubic light years of space

∠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.