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:
Answer: QN = 12
Step-by-step explanation: This quadrilateral is a paralelogram because its 2 opposite sides (NP and MQ) are parallel and the other 2 (MN and PQ) are congruent.
In paralelogram, diagonals bisect each other, which means QR = RN.
If QR = RN:
QR = 6
Then,
QN = QR + RN
QN = 6 + 6
QN = 12
<u>The diagonal QN of quadrilateral MNPQ is </u><u>QN = 12</u><u>.</u>
Answer:
The median
Step-by-step explanation:
The median always exists. The median does not have to be one of the data
Answer:
Step-by-step explanation:
(x+3)² -5 =0 , use the formula (a+b) ² = a²+b²+2ab
x²+9 +6x -5 =0 , combine like terms
x²+6x +4 =0, use the quadratic formula x = (-b±√b²-4ac)/2a
x= (-6 ±√6²-4*1*4)/2*1
x= (-6 ± √36-16) /2
x= (-6±√20)/2
x=(-6 +2√5)/2 and x=( -6-2√5) /2, factor 2 in the numerator and simplify
x= -3 +√5 and x= -3 -√5
