Aconditional relative frequency table is generated by row from a set of data. the conditional relative frequencies of the two categorical variables are then compared. if the relative frequencies are 0.48 and 0.52, which conclusion is most likely supported by the data? there is likely an association between the categorical variables because the relative frequencies are similar in value. an association cannot be determined between the categorical variables because the relative frequencies are similar in value. an association cannot be determined between the categorical variables because the relative frequencies are not similar in value. there is likely an association between the categorical variables because the relative frequencies are both close to 0.50.

the answer is b. An association cannot be determined between the categorical variables because the relative frequencies are similar in value.  

Step-by-step explanation:

∠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.

yes because they have the same amount of green and blur marbles.

step-by-step explanation: