Simplify 34x+13xy+12y

I WILL GIVE BRAINLIEST! DO NOT ANWSER IF YOU DON'T KNOW! Guy and Jim work at a furniture store . Guy is paid $185 per week plus

adoni [48]

Answer

X=18000

Step-by-step explanation:

7 0

3 years ago

the sum of ava’s and grace’s ages is 35. Ava is one year less than three time Grace’s age. How old is Ava, x and Grace, y expres

pantera1 [17]

Answer:

Ava is 26 and Grace age is 9

Step-by-step explanation:

x=ava's age

y=grace's age

3y-1=x

x+y=35

Substitute 3y-1 to x+y=35 for the value of x to find grace age

3y-1+y=35

4y-1=35

+1 +1

4y=36

y=9 now substitute 9, the y value in 3y-1=x to find ava's age

3(9)-1=x

x=26

to verify; 26+9=35

true 35=35

6 0

3 years ago

One third of the candidates for a job were 25 years old or younger. two sevenths of the candidates were at least 50 years old. i

Shalnov [3]

There are three groups of people: 25 years old and younger, between 25 to 50, and 50 years old and older. Their fractions must equal to 1 because these three together form the whole.

1 = 1/3 + 2/7 + x
where x is the fraction for people ages 25 to 50 years old
x = 8/21

So, the actual number of people ages 25 to 50 years old is:

84(8/21) = 32 people

7 0

3 years ago

What is the inverse of the function f(x) = 2x + 1?

hodyreva [135]

Answer:

The inverse f-1(x) = ( x - 2) / 2.

Step-by-step explanation:

Let 2x + 1 = y

Now find x in terms of y:

2x = y - 1

x = (y - 1) / 2

Replace the x by the inverse f-1(x) and the y by x, so we have:

f-1(x) = ( x - 2) / 2.

4 0

3 years ago

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.

3 0

3 years ago

Simplify 34x+13xy+12y​

Simplify 34x+13xy+12y