Which of the following is equivalent to 3x

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

4 years ago

You have 6 pints of glaze. It takes 7/8 of a pint to glaze a bowl and 9/16 of a pint to glaze a plate.

son4ous [18]

Answer:

a. How many bowls could you glaze?

6 bowls

How many plates could you glaze?

10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

How many plates can you glaze?

2 plates

How much glaze will be left over?

8/9 of glaze would be left

c. How many of each object could you glaze so that there is no glaze left over? Explain how you found your answer.

4 4/23 bowls and 4 4/23 plates

Step-by-step explanation:

We are told in the question

Number of pints of glaze = 6 pints

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

a.

How many bowls could you glaze?

7/8 pint = 1 bowl

6 pints = x

Cross Multiply

x = 6 pints/ 7/8

= 6 × 8/7

= 48/7

= 6.8571428571 bowls

We can only glaze whole numbers of a bowl, hence, we can only glaze 6 bowls

How many plates could you glaze?

9/16 pint = 1 bowl

6 pints = y

y = 6 pints/ 9/16

= 6 × 16/9

= 10.666666667 plates

We can only glaze whole number of a plate, hence, we can only glaze 10 plates

b. You want to glaze 5 bowls, and then use the rest for plates.

1 bowl = 7/8 pints

5 bowls = z

Cross Multiply

z = 5 × 7/8pints

= 4 3/8 pints of glaze would be used for bowls

How many plates can you glaze?

The rest is for plates, hence:

The amount of glaze left for plates is calculated as:

6 pints - 4 3/8

1 5/8 pints of glaze would be left over to glaze plates.

So therefore,

9/16 pints = 1 plate

1 5/8 pints =

= 2 8/9

Hence we can only glaze 2 plates

How much glaze will be left over?

2 8/9 pints - 2 pints

= 8/9 pints of glaze.

c. How many of each object could you glaze so that there is no glaze left over?

We have 6 pints of glaze

Number of pints to glaze a bowl= 7/8 of pint

Number of pints to glaze a plate = 9/16 of a pint

Let the number of each objects be represented by x

7/8 × x + 9/16 × x = 6 pints

= 4 4/23 bowls and 4 4/23 plates

3 0

3 years ago

I just need to know the third one pls

katen-ka-za [31]

Answer:

98

Step-by-step explanation:

we're doing PEDMAS, so do Exponents first.

3² = 9

8 + 10 x 9

Multiply

8 + 90

Add

98

4 0

3 years ago

sailfish are the fastest fish in the world they can swim 68 miles hour estimate how far sailfish can swim in 3 hours

nordsb [41]

68×3=204iles in 3 hour 204/3

5 0

4 years ago

Which ordered pair is a solution to the equation y= 6x - 2?

Dmitry [639]

Answer:

(5, 28)

Step-by-step explanation:

y= 6x - 2

Verifying pairs:

(1,0)

0 = 6*1 - 2
0 = 4 - incorrect

(5, 28)

28 = 6*5 - 2
28 = 30 -2
28 = 28 - correct

(6, 20)

20 = 6*6 - 2
20 = 36 - 2
20 = 34 - incorrect

(3, 19)

19 = 6*3 - 2
19 = 18 - 2
19 = 16 - incorrect

4 0

3 years ago

Which of the following is equivalent to 3x - 4y = 6?