When testing μ, how do we decide whether to use the standard normal distribution or a Student's t distribution? If the x distrib
1 answer:
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Answer:
500 boxes
Step-by-step explanation:
<em>hey there,</em>
<em />
< Since "p" stands for profits, change "p" to 0 (since the question says 0 profits).
Here is how your equation would look like:
Pretend like "n" is "x" since that is the variable we are trying to find here.
Solve for "n". Move all terms to the left side and set equal to zero. Then set each factor equal to zero.
"n" actually ends up equaling 500, -200.
Obviously, negative can't be the answer because you can't have a negative amount of boxes. So 500 boxes would be your answer. 200 CAN'T be your answer <em>either </em>because it is a negative!! >
<u>Hope this helped! Feel free to ask anything else.</u>
Answer:
a) P(X∩Y) = 0.2
b) = 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is: