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3 years ago

System of equations x-y-2z=4, -x+2y+z=1, x+y-3z=11. Solve by elimination

1 answer:

3 0

$x-y-2z=4 \\ -x+2y+z=1 \\ x+y-3z=11 \\ \\ \left \{ {{x-y-2z=4 } \atop {-x+2y+z=1}} \right \ and \ \left \{ {{-x+2y+z=1 } \atop {x+y-3z=11}} \right \\ \\ \hbox{1st system:} \\ x-y-2z=4 \\ \underline{-x+2y+z=1} \\ x-x+2y-y-2z+z=4+1 \\ y-z=5$

$\hbox{2nd system:} \\ -x+2y+z=1 \\ \underline{x+y-3z=11} \\ x-x+2y+y-3z+z=1+11 \\ 3y-2z=12 \\ \\ \hbox{now compare both:} \\ y-z=5 \ \ \ |\cdot (-2) \\ 3y-2z=12 \\ \\ -2y+2z=-10 \\ \underline{3y-2z=12 \ \ \ \ \ \ } \\ 3y-2y+2z-2z=12-10 \\ y=2 \\ \\ 2-z=5 \\ -z=3 \\ z=-3 \\ \\ x-2-2 \cdot (-3) = 4 \\ x-2+6=4 \\ x+4=4 \\ x=0$

The answer is: x=0, y=2, z=-3.

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f) The life of a power transmission tower is exponentially distributed, with mean life 25 years. If three towers, operated indep

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15.24% probability that at least 2 will still stand after 35 years

Step-by-step explanation:

To solve this question, we need to understand the binomial distribution and the exponential distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Probability of a single tower being standing after 35 years:

Single tower, so exponential.

Mean of 25 years, so $m = 25, \mu = \frac{1}{25} = 0.04$

We have to find $P(X > 35)$

$P(X > 35) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-0.04*35} = 0.2466$

What is the probability that at least 2 will still stand after 35 years?

Now binomial.

Each tower has a 0.2466 probability of being standing after 35 years, so $p = 0.2466$

3 towers, so $n = 3$

We have to find:

$P(X \geq 2) = P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.2466)^{2}.(0.7534)^{1} = 0.1374$

$P(X = 3) = C_{3,3}.(0.2466)^{3}.(0.7534)^{0} = 0.0150$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.1374 + 0.0150 = 0.1524$

15.24% probability that at least 2 will still stand after 35 years

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Help plz .................

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umm it's hardd kk it but I'll help

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