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3 years ago

12 divided 2 times 2 ^ 3 equals 120

Mathematics

2 answers:

Jlenok [28]3 years ago

6 0

12 divided 2 times 2 ^ 3 equals 48

Anna71 [15]3 years ago

3 0

What is the question here?

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-2/4 times 344/9=? Please help me . I dont really understand this question.

baherus [9]

Answer:

-19 1/9

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Is x = 6 a solution to the equation 5(x – 3) = x + 13?

deff fn [24]

Answer:

x=6 is not a solution

Step-by-step explanation:

We can substitute 6 as x to find out:

5(6-3)=6+13

first, because of PEMDAS we'll do 6-3

5(3)=6+13

multiply 5 by 3 and add 6 and 13

15=19

Since the statement is not true, that means that x=6 is not a solution

Hope this helps!

8 0

3 years ago

Ayuda 7×9÷2= note : btw how to get rid of this ad ?

JulijaS [17]

Step-by-step explanation:

7×9÷2

By using the rule of BODMAS , we get

7×9÷2

7×4.5

31.5

5 0

2 years ago

Read 2 more answers

Plz help.

zvonat [6]

$x-\ the price \ of \ the \ gift \ \\ \\ 10\% \ of \ x= \frac{10}{100}*x= \frac{1}{10}*x=\frac{x}{10}- the \ discount \\ \\x^{(10} -\frac{x}{10}= \frac{10x-x}{10}= \frac{9x}{10} -the \ new \ price \ \\ \\4*9.00 \$=36 \$ -the \ price \ after \ discount \\ \\ \frac{9x}{10}=36 =>x=\frac{10*36}{9} \\ \\ \boxed{x=40 \$}$

the variant A.$40.00

5 0

3 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