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mina [271]
3 years ago
9

Solve the system by elimination

Mathematics
1 answer:
Alja [10]3 years ago
3 0

First of all we will eliminate x from our equations. In order to do that we will use our first and second equation and then we will use second and third equation.

-2x+2y+3z=0...(1)\\-2x-y+z=-3....(2)

Upon subtracting 2 equation from 1 we will get,

3y+2z=3....(4)

Now we will use second and third equation to eliminate x.

-2x-y+z=-3....(2)\\2x+3y+3z=5....(3)

Adding 2nd and 3rd equation we will get,

2y+4z=2....(5)

Now we will find out y from our 4th and 5th equation.

2*(3y+2z)=2*3....(4)\\2y+4z=2....(5)

Upon subtracting 5th equation from 4th equation we will get,

4y=4\\y=1

Now let us find out z by substituting y's value in 5th equation.

2*1+4z=2\\4z=2-2\\z=0

Now we will find x from by substituting y and z's value in equation 1.

-2x+2*1+3*0=0\\-2x+2=0\\2=2x\\x=1

Therefore, x=1, y=1 and z=0 is the solution of the given system.

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Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

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(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

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F(200) = 1 - e^{-0.0143.200}

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = \sqrt{\frac{1}{\lambda^{2}} }

σ = \sqrt{\frac{1}{0.0143^{2}} }

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - e^{-0.0143*209.79})

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

1 - e^{-\lambda.m} = 0.5

-e^{-\lambda.m} = - 0.5

ln(e^{-0.0143.m}) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0
3 years ago
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