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Answer:
39.10%
Step-by-step explanation:
The given timeout distributions are:
Exp (1/10), Exp (1/12)
The equation to use is the following:
F (x) = 1 - e ^ (- a / b); x> 0
So the probability is:
P (food has not arrived after waiting 25 minutes | R) = e ^ (- 10/10) = 0.3679
P (food has not arrived after waiting 25 minutes | S) = e ^ (- 12/15) = 0.2865
So:
P (food has not arrived after waiting 25 minutes) = (food has not arrived after waiting 25 minutes | R) * P (R) + (food has not arrived after waiting 25 minutes | S) * P (S)
Replacing:
P (food has not arrived after waiting 25 minutes) = 0.3679 * 1/3 + 0.2865 * 2/3 = 0.3136
However:
P (R | food has not arrived after waiting 25 minutes) = (food has not arrived after waiting 25 minutes | R) * P (R) / P (food has not arrived after waiting 25 minutes)
P (R | food has not arrived after waiting 25 minutes) = 0.3679 * 1/3 / 0.3136 = 0.3910
It means that the probability is 39.10%
The right answer for the question that is being asked and shown above is that: "B.second quarter." Use the card collection box-and-whisker plot to solve.<span> </span><span>In the second quarter that the data are at least concentrated.</span>
I don't know. I'll have to figure it out.
Probability =
(number of successful possible results) / (total possible results) .
Total possible results for the spinner . . . 8
Successful ones (getting a 1, 2, 3, 4, or 5) . . . 5
Theoretical probability = 5/8 .
So you expect (5/8 of 150) spins to be successful.
(5/8 times 150) = 93.75
Yes !
93 or 94 is a good answer.
