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galben [10]
3 years ago
5

The libary has at least 5,000 books witch inequality represent the number of books,b,at the library

Mathematics
1 answer:
Mazyrski [523]3 years ago
7 0
B\geq 5,000

because the library has more than or = to 5000 books.
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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
There are a total of 375 flamingos, parrots, herons, and chachalacas at the Famous Bird Sanctuary. The ratio of flamingos to par
Tems11 [23]

Answer:

75

Step-by-step explanation:

Since the ratio between flamingos and chachalacas is 2:1, we see that the ratio of flamingos to chachalacas to parrots to herons is 10:5:7:3. The sum of these numbers is 25, so every 'ratio number' gets 375/25 = 15 animals. So the number of chachalacas is 15 * 5 = 75.

3 0
3 years ago
5(3a-1)-2(3a-2)=3(a+2)+v
STatiana [176]

Answer:

<u>The answer is option C. 6a-7</u>

Step-by-step explanation:

Given that

5(3a-1)-2(3a-2)=3(a+2)+v

Solve for v

∴ v = 5(3a-1)-2(3a-2) - 3(a+2)

∴ v = 15a - 5 - 6a + 4 - 3a - 6

∴ v = 15a - 6a - 3a - 5 + 4 - 6

∴ v = 6a - 7

<u>So the answer is option C. 6a-7</u>

5 0
3 years ago
Ahhh help I have no idea
Monica [59]

Answer:

Solve 3-6x >= 0

6x <= 3

x <= 1/2

Domain: All Real Numbers <= 1/2

7 0
3 years ago
Options:
AlexFokin [52]
Oh idk and idc cuz i just really don’t so yea
5 0
3 years ago
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