According to a survey, about 30% of people ages 18 and older in a city have a cat. You ask 5 randomly chosen people (ages 18 and older) whether they have a cat. What is the probability that at least 2 people have a cat? Round your answer to the nearest hundredth.

Step-by-step explanation:

When they ask what the prob that at most 3/4 people asked have a pet, they are asking what is the prob that 0, 1, 2, or 3 people answer yes. So you could calculate all 4 of those and then add them together.

However, in this case, since there's only 1 other possible outcome (all 4 have a pet), we could just calculate the prob of that and subtract it from 1 to get the prob of everything else happening.

We'll use the binomial distribution formula where n=4, k=4, and P=.6.

nCk * P^n * (1-P)^(n-k)4C4 * (.6)^4 * (.4)^0.1296

So the prob of at least 3 having pets = 1 - P(all 4 have pets) which is 1-.1296 = .8704.  You could double check this by calculating the prob of the other 4 outcomes and then adding them together.

∠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.