Question 8(Multiple Choice Worth 1 points) (06.05 MC)
A paper cup is dropped and its landing position is recorded. The cup can land on the side, on the open end, or on the closed end. The results of 20 trials are shown in the table below:
Paper Cup Experiment
# of times occurred
Open
HT III
Closed
Side
HT III
Based on the table, which of the following best compares the experimental probability of the cup landing on its open end with the experimental probability of the cup landing on its closed end?
The probabilities are equal.
The probability of landing on the open end is greater.
The probability of landing on the closed end is greater.
O No conclusion can be made.

Table Given in the question :

Open = HT 111 = 8

Side = 1111 = 4

Side = HT 111 = 8

The probability of landing on the open end is greater.

Step-by-step explanation:

Given the experimental probability distribution :

Open = HT 111 = 8

Side = 1111 = 4

Side = HT 111 = 8

The experimental probability is the ratio of the number of times an event occurs and the total number of trials.

P(A) = number of times A occurs / total number of trials

Where A is defined event.

COMPARING the probabilities of open and closed events

P(open) = 8 / 20 = 2 / 5

P(closed) = 4 / 20 = 1/5

2/5 > 1/5

P(open) > P(closed)

answer: what if i dont want to show my work boi.

step-by-step explanation: im watching dbz and eating popcorn leave meh alone

∠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.