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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

Step2247 [10]

Answer:

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

4 0

3 years ago

What can you predict regarding the probability of the coin landing heads up?

Kisachek [45]

Answer:

the probability is ½

Step-by-step explanation:

think so

3 0

3 years ago

Pls answer asap 7 - 3 r = r - 4 ( 2 + r)

Marianna [84]

Answer:

no real answer

Step-by-step explanation:

distribute the -4 and combine all of the like terms on the left side = r-8-4r, then -8-3r

then we have 7-3r=-8-3r

from here, we can already tell that there's no real answer. this is because the two -3r will cancel, leaving no variable.

since 7 doesn't equal -8, there is no answer.

if, for example, the value on both sides of the equal sign were the same after the variable was eliminated, then your answer would be all real numbers

8 0

3 years ago

A man 50 years old has 8 sons born of equal intervals. The sum of the ages of the father and sons is 186. What is the age of the

Rudiy27

<span>let x be the interval, then: 186 = 50 + 3 + (3+x) + (3+2x) + (3+3x) + (3+4x) + (3+5x) + (3+6x) + (3+7x) 186 = 74 + 28x x = 4 Eldest son age = 3+7x = 3+28 = 31.</span>

8 0

4 years ago

(2*)2-3×2*+2=0<br>4m-15°(×)m+75°

Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

0

5

−

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

2

and \displaystyle {R}_{3}R

3

.

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

1

−

12

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

1

−

12

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

0

5

−

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

2

and \displaystyle {R}_{3}R

3

.

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

1

−

12

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

0

1

−

12

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0

2 years ago