Evaluate the following for f(x) = 3(2)^x 1. f (2) = 2. f (-1) = 3. f (10) =

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

Suppose f and g are continuous functions such that g(2) = 6 and lim x → 2 [3f(x) + f(x)g(x)] = 36. find f(2).

True [87]

Answer: f(2) = 4

Step-by-step explanation:

F(x) and g(x) are said to be continuous functions

Lim x=2 [3f(x) + f(x)g(x)] = 36

g(x) = 2

Limit x=2

[3f(2) + f(2)g(2)] = 36

[3f(2) + f(2) . 6] = 36

[3f(2) + 6f(2)] = 36

9f(2) = 36

Divide both sides by 9

f(2) = 36/9

f(2) = 4

7 0

3 years ago

The LCD of 2/5, 3/2, and 1/2 is

soldier1979 [14.2K]

Answer:

10

Step-by-step explanation:

the lowest number that 5 and 2 go into is 10 a quick way to find LCD is just to multiply the denominators and that just see if there is any smaller numbers

7 0

3 years ago

Determine the slope of a line that goes through (2,7) and (11,10)

tia_tia [17]

Answer:

1/3

Use the slope formula and simply the fraction.

4 0

3 years ago

A scuba diver descends farther down into the ocean from an initial depth of 15.2 feet below sea level. The scuba diver descends

gavmur [86]

Answer:

r <= 12.6

Step-by-step explanation:

Let's calculate the rate at which he descends:

r = (46.7 - 15.2)/212 = 0.15 feet/minute

5 0

2 years ago

Evaluate the following for f(x) = 3(2)^x1. f (2) =2. f (-1) =3. f (10) =​

Evaluate the following for f(x) = 3(2)^x1. f (2) =2. f (-1) =3. f (10) =