Find c.
1 answer:
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Answer:
(A) The probability that a randomly selected adult is either overweight or obese is 0.688.
(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C) The events "overweight" and "obese" exhaustive.
(D) The events "overweight" and "obese" mutually exclusive.
Step-by-step explanation:
Denote the events as follows:
<em>X</em> = a person is overweight
<em>Y</em> = a person is obese.
The information provided is:
A person is overweight if they have BMI 25 or more but below 30.
A person is obese if they have BMI 30 or more.
P (X) = 0.331
P (Y) = 0.357
(A)
The events of a person being overweight or obese cannot occur together.
Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.
So, P (X ∩ Y) = 0.
Compute the probability that a randomly selected adult is either overweight or obese as follows:
Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.
(B)
Commute the probability that a randomly selected adult is neither overweight nor obese as follows:
Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C)
If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.
For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.
In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.
Thus, the events "overweight" and "obese" exhaustive.
(D)
Mutually exclusive events are those events that cannot occur at the same time.
The events of a person being overweight and obese are mutually exclusive.
Answer:
Step-by-step explanation:
To evaluate :
Solution:
Two negatives multiply to become a positive.
Thus, we can remove parenthesis by reversing the signs of the fraction by multiplying the negative outside.
⇒
Since the denominators are same for both fractions, so we simply add the numerators.
⇒
⇒ (Answer)