Select each of the following experiments that are binomial experiments:

a) Drawing 3 balls without replacement from a box that contains 10 balls, 4 of which are red and 6 are blue and observing the colors of the drawn balls.
b) Drawing 3 balls with replacement from a box that contains 10 balls, 4 of which are red and 6 are blue and observing the colors of the drawn balls.
c) Selecting a few households from New York City and observing whether or not they own stocks when it is known that 40 % of all households in New York City own stocks.

a) No.

b) Yes.

c) Yes.

Step-by-step explanation:

a) No.

As being without replacement, the probabilities of each color in each draw change depending on the previous draws.

This is best modeled by an hypergeometric distribution.

b) Yes.

As being with replacement, the probabilities for each color is constant.

Also, there are only two colors, so the "success", with probability p, can be associated with the color red, and the "failure", with probability (1-p), with the color blue, for example.

(With more than two colors, it should be "red" and "not red", allowing only two possibilities).

c) Yes.

The answer is binary (Yes or No) and the probabilities are constant, so it can be represented as a binomial experiment.

the ratio of white tiles to blue tiles is 48: 16. the ratio in it's simplest form is 3: 1.

step-by-step explanation:

ratio: take the first number and add a colon next to it, then add the second number after that. make sure the ratio order matches up with the written problem!

simplest form: we can simplify the ratio 48 : 16 by dividing both terms by the greatest common factor (gcf).

the gcf of 48 and 16 is 16.

divide both terms by 16:

48 ÷ 16 = 3

16 ÷ 16 = 1

reducing to simplest form:

48 : 16 = 3 : 1

∠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.