Answer:
(c) Observational Distribution
(d) Diet Choice
Step-by-step explanation:
Out of 1033 American adults interviewed in July 2018, 5% consider themselves to be vegetarian.
Since the poll observes the diet habit of the respondents, the cited 5% is part of the Observational distribution of whether the respondent is a vegetarian or not (which is the Diet Choice).
If on the other hand, the poll seeks to manipulate the conditions of the study, it would have been an experimental distribution.
Answer:

Step-by-step explanation:
We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.
Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.
Since the probability of two independent compound events is always the product of probabilities of both events.

Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.

Probability of picking blue ball out of 16 (4+7+5) marbles:

Now let us find probability of Monica picking a red and then a blue marble.





Therefore, the probability of picking a red and then blue marble is
.
The parent function is f(x) = x^3
The domain are all x values (-infinity, infinity)
The range are all y values (-infinity, infinity)
No I don’t think it’s possible
Answer: -1.51
Step-by-step explanation:
Let
be the population mean rating of car.
As per given , we have


Since population variance is known, so we use z-test.
Test statistic : 
, where n= sample size
= Sample mean
= variance
As per given ,
n=300



Put these values in formula we get



Hence, the value of the test statistic is z= -1.51 .