An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
1 answer:
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Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) =
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);
e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
(c) The Poisson distribution is presented as follows;
p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4) .
Answer:
the minimum expected elastic modulus is 372.27 Gpa
Explanation:
First we put down the data in the given question;
Volume fraction = 0.84
Volume fraction of matrix material = 1 - 0.84 = 0.16
Elastic module of particle = 682 GPa
Elastic module of matrix material = 110 GPa
Now, the minimum expected elastic modulus will be;
= (
×
) / (
+
)
so we substitute in our values
= (682 × 110 ) / ( [ 682 × 0.16 ] + [ 110 × 0.84] )
= ( 75,020 ) / ( 109.12 + 92.4 )
= 75,020 / 201.52
= 372.27 Gpa
Therefore, the minimum expected elastic modulus is 372.27 Gpa