Mathematics, 09.07.2019 08:10 ferg6

(b) what is the probability that more than eleven loads occur during a 4-year period

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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:
$\frac{4}{0.4} =10 \ loads$

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

$P(k \ events \ in \ an \ interval)= \frac{\lambda^ke^{-\lambda}}{k!}$

where: k = 0, 1, 2, . . ., 11 and λ = 10.

$P(k=0)= \frac{10^0e^{-10}}{0!} =0.000045 \\ \\ P(k=1)= \frac{10^1e^{-10}}{1!} =0.000454 \\ \\ P(k=2)= \frac{10^2e^{-10}}{2!} =0.002270 \\ \\ P(k=3)= \frac{10^3e^{-10}}{3!} =0.007567 \\ \\ P(k=4)= \frac{10^4e^{-10}}{4!} =0.018917 \\ \\ P(k=5)= \frac{10^5e^{-10}}{5!} =0.037833 \\ \\ P(k=6)= \frac{10^6e^{-10}}{6!} =0.063055 \\ \\ P(k=7)= \frac{10^7e^{-10}}{7!} =0.090079 \\ \\ P(k=8)= \frac{10^8e^{-10}}{8!} =0.112599 \\ \\ P(k=9)= \frac{10^9e^{-10}}{9!} =0.125110$

$P(k=10)= \frac{10^{10}e^{-10}}{10!} =0.125110 \\ \\ P(k=11)= \frac{10^{11}e^{-10}}{11!} =0.113736$

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 0.4283

Part 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:

$\frac{\lambda^0e^{-\lambda}}{0!} \leq0.3 \\ \\ \Rightarrow e^{-\lambda}\leq0.3 \\ \\ \Rightarrow-\lambda\leq\ln{0.3} \\ \\ \Rightarrow -\lambda\leq-1.204 \\ \\ \Rightarrow\lambda\geq1.204$

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:

$\frac{4}{1.204} =3.3 \ years$

Answer from: Quest

- 2 $x^{4}$ + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 $x^{4}$ + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 $x^{4}$ + 10x³ - 21x² + 19x - 4 ← in standard form

Answer from: Quest

step-by-step explanation:

if it takes 8 pounds to make 12 divide 12 by 8 and you get 1.5, 1.5 x 16 = 24

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