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3 years ago

How to solve 2^2-5t+12=0

2 answers:

5 0

2^2 - 5t + 12 = 0

4 - 5t + 12 = 0

-5t + 16 = 0

-5t = -16

t = 16/5

Answer: 16/5 in fraction form or 3.2 in decimal form

Solnce55 [7]3 years ago

5 0

2² - 5t + 12 = 0

So, because we are solving for t, you must get t isolated! :)

Simplify the equation.

4 - 5t + 12 = 0

Add like terms.

16 - 5t = 0

Subtract 16 from both sides.

-5t = 0 - 16

Simplify.

-5t = -16

Divide both sides by -5.

t = 16/5 OR 3 1/5

~Hope I helped!~

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Answer:

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Step-by-step explanation:

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Read 2 more answers

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

I attached the math problem below and I hope u guys help me. Thxs to who's gonna answer!

vova2212 [387]

Answer:

Numbers | least to greatest | opposites | opposites least to greatest

1/4, -1/2 | -1/2, 1/4 | -1/4, 1/2 | -1/4, 1/2

2, -10 | -10, 2 | -2, 10 | -2, 10

0, 3 1/2 | 0, 3 1/2 | 0, -3 1/2 | -3 1/2, 0

-5, -5.6 | -5.6, -5 | 5, 5.6 | 5, 5.6

24 1/2, 24 | 24, 24 1/2 | -24 1/2 -24 | -24 1/2 -24

-99.9, -100 | -100, -99.9 | 99.9, 100 | 99.9, 100

-0.05, -0.5 | -0.5, -0.05 | 0.5, 0.05 | 0.05, 0.5

-0.7, 0 | -0.7, 0 | 0.7, 0 | 0, 0.7

100.02, 100.04 | 100.02, 100.04 | -100.02, -100.04 | -100.04, -100.02

Hope this helps you!!!!

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2 years ago