The owner of a local movie theater keeps track of the number of tickets sold in each purchase. The owner determines the probabilities based on these records. Let X represent the number of tickets bought in one purchase. The distribution for X is given in the table. A 2-column table with 5 rows. Column 1 is labeled number of tickets with entries 1, 2, 3, 4, 5. Column 2 is labeled Probability with entries 0.17, 0.55, 0.20, 0.06, 0.02.

Which of the following correctly represents the probability of a randomly selected purchase having at most three tickets?

Which of the following correctly represents the probability of a randomly selected purchase having at most three tickets?
(not d)
P(X ≤ 3)
P(X < 3)
P(X ≥ 3)
P(X > 3)

step-by-step explanation:

just plug in each number for x and see which ones work and don't

∠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.

he has now walked a total of what?

if i know what this is maybe i can

step-by-step explanation: