A grocer noticed that customers buy both pet and baby supplies with a probability of 0.12. Considered separately, the probability that customer buys pet supplies is P(A) = 0.64, and the probability that a customer buys baby supplies is P(B) = 0.25.

Move expressions and numbers into the blanks to create an equation that represents the probability that a customer buys either pet

supplies or baby supplies.

0.77

Step-by-step explanation:

Given that:

P(A) = 0.64

P(B) = 0.25

P(AnB) = 0.12

Probability that customer buys either pet or baby supplies ; P(AuB)

P(AuB) = P(A) + p(B) - p(Ann)

P(AuB) = 0.64 + 0.25 - 0.12

P(AuB) = 0.77

∠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.

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step-by-step explanation: