Answer:
Correct me if I’m wrong, but the answer should be f(t) = 2 cos pi over 6 t + 4.
Step-by-step explanation:
(X+5)²=4/4
X+5=1, X+5=-1
X=1-5 X=-1-5
X=-4 X=-6
Answer:
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- <u><em>Yes, it is reasonable to expect that more than one subject will experience headaches</em></u>
Explanation:
Notice that where it says "assume that 55 subjects are randomly selected ..." there is a typo. The correct statement is "assume that 5 subjects are randomly selected ..."
You are given the table with the probability distribution, assuming, correctly, the binomial distribution with n = 5 and p = 0.732.
- p = 0.732 is the probability of success (an individual experiences headaches).
- n = 5 is the number of trials (number of subjects in the sample).
The meaning of the table of the distribution probability is:
The probability that 0 subjects experience headaches is 0.0014; the probability that 1 subject experience headaches is 0.0189, and so on.
To answer whether it <em>is reasonable to expect that more than one subject will experience headaches</em>, you must find the probability that:
- X = 2 or X = 3 or X = 4 or X = 5
That is:
- P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).
That is also the complement of P(X = 0) or P(X = 1)
From the table:
- P(X = 0) = 0.0014
- P(X = 1) = 0.0189
Hence:
- 1 - P(X = 0) - P(X = 1) = 1 - 0.0014 - 0.0189 = 0.9797
That is very close to 1; thus, it is highly likely that more than 1 subject will experience headaches.
In conclusion, <em>yes, it is reasonable to expect that more than one subject will experience headaches</em>
The answer is the option D.
You have to use implicit derivative
1) dy / dx = y'
2)
