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Georgia [21]
3 years ago
5

Quien me ayuda (((2) ³) ²) ⁰ ((((-3) ²) ²) ³) ⁰

Mathematics
1 answer:
blagie [28]3 years ago
3 0

Answer:

la respuesta a ambos es 1 porque si elevas cualquier numero a un exponente de 0 este queda en 1

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Multiply and combine like terms:<br> -5x – 4(9 – 2x)
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-5x-36+8x

3x-36

3(x-18)

3 0
2 years ago
The Pyraminx is a Rubik's cube-type toy in the shape of a tetrahedron. The Pyraminx shown below has edges 15\,\text{cm}15cm long
____ [38]

Answer:

8.727

Step-by-step explanation:

Since the below edge, d = 15 cm, and the height, h = 12.2 cm, we will use the Pythagorean  theorem to find the distance r.

d^2=r^2+h^2

(15)^2 = r^2+(12.2)^2

r^2 = (15)^2-(12.2)^2=225-148.84=76.16

r=\sqrt{76.16}=8.727

4 0
2 years ago
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
Statistics: Describing Histograms
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Given the polynomial expression 3x^2 + 3bx - 6x - 6b, factor completely.
My name is Ann [436]
The answer is 3 (x+b) (x-2)



Yw! :)
3 0
2 years ago
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