(b) y/3 + y/4 + y/5 = y/3 + y/4 + y/5 + 1
(c) z/3 + z/4 + z/5 = z/4 + z/5 + z/6 + 1$
I. Probability that Dave is a rollerblader = 50%
II. Probability that Aimee does not have kids = 75%
III. Probability that Mitch is a rollerblader = 80.65%
The required, none of the statements are correct, as of the given conditions.
Probability can be defined as the ratio of favorable outcomes to the total number of events.
Here,
We are given a survey of a random sample of residents in a neighborhood who have kids and who also rollerblade. Let's denote the events as follows:
R: the event that a resident rollerblades
K: the event that a resident has kids
Based on the information given in the problem, we know that:
Dave does not have kids, so P(K) = 0 for Dave. Therefore, the probability that Dave is a rollerblader is 0, not 50%. So statement I is incorrect.
Aimee has never rollerbladed, so P(R) = 0 for Aimee. We don't know anything about whether Aimee has kids or not, so we cannot determine the probability that she does not have kids. So statement II is incorrect.
We don't have any information about whether Mitch has kids or not, so we cannot determine the probability that Mitch is a rollerblader. Therefore, statement III is also incorrect.
Therefore, none of the statements are correct.
Learn more about probability here:
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Answer:
1. dave has a low chance of rollerblading if he is focusing on his career.
Step-by-step explanation:
14³/₈ or 14.375 pounds.
Given:
The vet tells him his cat's weight is ⁵/₈ as his dog's weight.
Question:
How much does his cat weigh?
The Process:
Since the vet tells him his cat's weight is part of his dog's weight by ⁵/₈ part, let us create an 8-unit pattern that represents the dog's weight.
And now, let us find out the cat's weight. As we know, the cats' weight is ⁵/₈ the dog's weight.
The cat's weight:
Recall this
Recall this
as a mixed fraction
as a decimal fraction
Thus, his cat weigh is
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