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2 years ago

8

Use the Poisson model to approximate the probability. Round your answer to four decimal places. Suppose the probability of a maj

or earthquake on a given day is 1 out of 15,000. Use the Poisson distribution to approximate the probability that there will be at least one major earthquake in the next 2000 days.

1 answer:

fomenos2 years ago

7 0

The value of lambda for 2000 days = 2000 * (1/15000) = 2/15.
$P(X=1)=\frac{e^{-\lambda}\lambda^{x}}{x!}=0.1167$

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A water dispenser has a capacity of 20,000 milliliters. Use the facts to find how much this is in liters .

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The water dispenser's capacity in liters is 20liters

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Given:

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let the capacity of the water dispenser in Liters = y

$\begin{gathered} 1000ml\text{ = 1 L} \\ 20000ml\text{ = y} \\ cross\text{ multiply:} \\ y(1000)\text{ = 1\lparen20000\rparen} \\ 1000y\text{ = 20000} \end{gathered}$ $\begin{gathered} divide\text{ both sides by 1000:} \\ y\text{ = }\frac{20000}{1000} \\ y\text{ = 20} \end{gathered}$

The water dispenser's capacity in liters is 20liters

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1 year ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

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Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

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$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

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Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

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