Organisms are present in ballast water discharged

Ответ:

TrueKing184

12.01.2022

Mathematics

a) The probaility is 0.333.

b) The probability is 0.125.

c) The volume is 0.512 m3.

Step-by-step explanation:

Organisms are present in ballast water discharged with a concentration of 10 organisms/m3.

That is our rate of the Poisson process

$\lambda=10\,org/m^3$

a) The probability of having at least 8 organisms is equal to the sum of the probabilities of having from 0 to 8 organisms:

$P(k\leq8)=\sum_{i=0}^{8}P(k=i)=\sum_{i=0}^{8}\frac{\lambda V^i e^{-\lambda V}}{i!} \\\\P(k\leq8)=0.000+0.000+0.002+0.008+0.019+0.038+0.063+0.090+0.113\\\\P(k\leq8)=0.333$

Note: In this case, the volume V is 1 m3.

b) In this case, the volume is 1.5m3 so we have to multiply the rate by 1.5. Then it becomes:

$\lambda V=10\,org/m^3\cdot1.5m^3=15org$

The standard deviation of this distribution is

$\sigma=\sqrt{\lambda}=\sqrt{15}\approx 3.873$

We have to calculate the probability of exceeding 19 organisms in 1.5m3:

$k=15+3.873 \approx19$

We have that the probability of having more than 19 org. is equal to one substracting the probability of having equal or less than 19 org:

$P(k19)=1-P(k\leq19)\\\\P(k19)=1-\sum_{i=0}^{19}P(k=i)=1-\sum_{i=0}^{19}\frac{\lambda V^i e^{-\lambda V}}{i!} \\\\P(k19)=1-(0.000+0.000+0.000+0.000+0.001+0.002+0.005+0.010+0.019+0.032+0.049+0.066+0.083+0.096+0.102+0.102+0.096+0.085+0.071+0.056)\\\\P(k19)=1-0.875=0.125$

c) We have to calculate the volume such that there is a probability P=0.994 of having at least one organism in the water. This can be calculated as one less the probability of having zero organisms.

$P(x\leq1)=1-P(0)=1-\frac{\lambda V^0 e^{-\lambda V}}{0!} \\\\P((x\leq1)=1-\frac{1\cdot e^{-10 V}}{1}=1-e^{-10V}=0.994\\\\e^{-10V}=1-0.994=0.006\\\\-10V=ln(0.006)=-5.116 \\\\V=5.116/10=0.512$

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Ответ:

spazzinchicago

09.03.2020

Mathematics

Step-by-step explanation:

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(

)(x)=

g(x)

f(x)

= \small{\dfrac{3x+2}{4-5x}}=

4−5x

3x+2

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