Let x denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. suppose that for banner-tailed kangaroo rats, x has an exponential distribution with parameter λ = 0.01352. (a) what is the probability that the distance is at most 100 m? at most 200 m? between 100 and 200 m? (round your answers to four decimal places.)

Answer with Step-by-step explanation:

We are given that

X=Distance (m) that animal moves from its birth site to the first territorial vacancy it encounters.

Parameter=

Exponential distribution of probability is given by

Cumulative distribution is given by

Where

a.We have to find the probability that the distance is at most 100m.

Hence, the probability that the distance is at most 100 m=0.7413

b.

Hence, the probability that the distance is at most 200m =0.9331

c.

Hence, the probability that the distance between 100 m and 200 m=0.1918

answer: its because of life

step-by-step explanation: i dont know

∠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.