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3 years ago

Use the elimination method to solve the system of equations.Choose the correct ordered pair. - x + y = - 2 and 2 x - 3y = 4

Mathematics

1 answer:

Zolol [24]3 years ago

8 0

Answer:(2,0) x=2,y=0

Step-by-step explanation:

Let's solve your system by elimination.

−x+y=−2;2x−3y=4

Multiply the first equation by 2,and multiply the second equation by 1.

2(−x+y=−2)

1(2x−3y=4)

Becomes:

−2x+2y=−4

2x−3y=4

Add these equations to eliminate x:

−y=0

Then solve−y=0 for y:

−y=0

−y/−1 =0/−1

(Divide both sides by -1)

y=0

Now that we've found y let's plug it back in to solve for x.

Write down an original equation:

−x+y=−2

Substitute0foryin−x+y=−2:

−x+0=−2

−x=−2(Simplify both sides of the equation)

−x/−1 = −2/−1

(Divide both sides by -1)

x=2

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Find the area of the composite figure.<br> I need a quick answer on pleas

Maru [420]

Answer:

Area if rectangle= 55

Area of triangle=30

Total area=85

11x5=55

23-11=12

5/2=2.5

2.5x12=30

55+30=85

8 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

4 years ago

Candace has $20 more than Amar. Their combined money totals $95. Write a system of equations represents this situation

STALIN [3.7K]

Answer:

Candace has $57.5 and Amar has $37.5.

Step-by-step explanation:

x+(x+20)=z

x+20=y

x+y=z

--------------

2x+20=95

2x=95-20

2x=75

x=75/2

x=37.5

37.5+20=y=57.5

5 0

3 years ago

Read 2 more answers

Complete the equation of the line whose y-intercept is (0,5) and the slope is 9.

madreJ [45]

Answer:

9x+y−5=0 9 x + y − 5 = 0 .

Step-by-step explanation:

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