Answer:
Given that an article suggests
that a Poisson process can be used to represent the occurrence of
structural loads over time. Suppose the mean time between occurrences of
loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a
4-year period? c). How long must a time period be so that the probability of no loads
occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a
4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads
occurring during that period is at most 0.3 is given by: 3.3 years
Step-by-step explanation:
F(x) = (x + 1)(x - 2)
f(x) = x(x - 2) + 1(x - 2)
f(x) = x(x) - x(2) + 1(x) - 1(2)
f(x) = x² - 2x + x - 2
f(x) = x² - x - 2
The graph is misleading because the year’s interval is not constant.
The first year to the second year, the gap is 1 year; in the second to the
third year, the gap is 2; in the third to the fourth year is 4; and the fourth
to the fifth year is 6.
You have 180 tablets right? You also have to take it three times a day.
180 pills /3 pills a day = <u>60 days</u>
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C=pi(d)
So d=7ft
Then pi(7ft)=21.98
