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il63 [147K]
3 years ago
13

A system has two components: A and B. The operating times until failure of the two components are independent and exponentially

distributed random variables with parameter 2 for component A, and parameter 3 for component B. The system fails at the first component failure.
(a) - Read section 1.5.2 in the textbook.
(b) - What is the mean time to failure for component A and for component B
Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

E(x) = \frac{1}{2} -- Component A

E(x) = \frac{1}{3} -- Component B

Step-by-step explanation:

Given

Distribution = Exponential

\lambda = 2 --- Component A

\lambda = 3 --- Component B

Solving (a): The mean time of A

The mean of an exponential distribution is:

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

We have:

\lambda = 2 --- Component A

E(x) = \frac{1}{2}

Solving (b): The mean time of B

The mean of an exponential distribution is:

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

We have:

\lambda = 3 --- Component B

E(x) = \frac{1}{3}

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Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots StartRoo
anastassius [24]

Answer:

a) f(x)=3x^{3}-6x^{2}-15x+30

Step-by-step explanation:

1) In this question we've been given "a", the leading coefficient. and two roots:

x_{1}=\sqrt{5}\:x_{2}=2

2) There's a theorem, called the Irrational Theorem Root that states:

If one root is in this form x'=\sqrt{a}+b  then its conjugate x''=\sqrt{a}-b. is also a root of this polynomial.

Therefore

x_3=-\sqrt{5}

3) So, applying this Theorem we can rewrite the equation, by factoring. Remembering that x is the root. Since the question wants it in this expanded form then:

f(x)=3(x-\sqrt{5})(x+\sqrt{5})(x-2)\Rightarrow 3x^{3}-6x^{2}-15x+30

4 0
2 years ago
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chubhunter [2.5K]
(5y-2)to the power of 2
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3 years ago
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A student draws a quadrilateral that has two sides measuring 5 cm and two sides measuring 8 cm, and the sides opposite to each o
ch4aika [34]

Answer:

B

Step-by-step explanation:

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3 0
2 years ago
The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation of 1.4. The distributio
posledela

Answer:

Approximate probability that \overline{X} is less than 2 = 0.1515

Step-by-step explanation:

Given -

Mean (\nu ) = 2.2

Standard deviation (\sigma  ) = 1.4

Sample size ( n ) = 52

Let  \overline{X} be the mean of accidents per week at the intersection during one year (52 weeks) .

probability that \overline{X} is less than 2 =

P(\overline{X}< 2)  = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}< \frac{2 - 2.2 }{\frac{1.4}{\sqrt{52}}})   Putting (Z =\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}})

                 = P(Z< - 1.03)    ( Using Z table )

                 = 0.1515

7 0
2 years ago
15. Simplify an expression for the area of the<br>rectangle shown below.<br>8x - 2.<br>4.<br>​
lyudmila [28]

Answer:

Area of the rectangle  A   = 32 x - 8

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that the length of the rectangle(l) = 8x-2

Given that the width of the rectangle (w) = 4

Area of the rectangle = length × width

                         A   = (8x - 2) × 4

                         A   = 32 x - 8

<u><em>Final answer</em></u>:-

Area of the rectangle  A   = 32 x - 8

3 0
3 years ago
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