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2 years ago

Help please. only answer if you know :(

Mathematics

1 answer:

Harrizon [31]2 years ago

7 0

Answer:

Step-by-step explanation:

<u>Right Triangles</u>

In right triangles, where one of its internal angles measures 90°, the trigonometric ratios are satisfied.

We have completed the figure below with the missing internal angle A that measures A = 90° - 29° = 61° because the lines marked with an arrow are parallel.

Given the internal angle A, we can relate the unknown side of length x with the known side length of 500 ft, the hypotenuse of the triangle. We use the sine ratio:

$\displaystyle \sin 61^\circ=\frac{\text{opposite leg}}{\text{hypotenuse}}$

$\displaystyle \sin 61^\circ=\frac{x}{500}$

Solving for x:

$x = 500 \sin 61^\circ$

Calculating:

x = 437.3 ft

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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.

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3 years ago

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En una empresa trabajan 60 personas. Usan gafas el 16% de los hombres y el 20% de las mujeres. Si el número total de personas qu

DIA [1.3K]

Pregunta completa:

En una empresa trabajan 60 personas. Usan gafas el 16% de los hombres y el 20% de las mujeres. Si el numero total de personas que usan gafas es 11. ¿Cuantos hombres y mujeres hay en la empresa?

Responder:

Hombres = 25

Mujeres = 35

Explicación paso a paso:

Dado lo siguiente:

Número de personas que trabajan en la empresa = 60

Porcentaje de hombres que usan anteojos = 16%

Porcentaje de mujeres que usan anteojos = 20%

Número total de personas que usan anteojos = 11

Suponga, Número de hombres en la empresa = m

Número de mujeres = número total - número de hombres = 60 - m

Por lo tanto,

16% de los hombres = 0,16 m

20% de mujeres = 0,2 (60 - m) = 12 - 0,2 m

Por lo tanto,

0,16 m + 12 - 0,2 m = 11

- 0,04 m = 11 - 12

-0,04 m = - 1

m = 1 / 0.04 = 25

Por tanto, Número de hombres en la empresa = m = 25

Número de mujeres en la empresa = (60 - m) = (60 - 25) = 35 mujeres

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3 years ago