Answer:
The answer is "Both Technician A and Technician B".
Explanation:
The cylinder Testing is intended to assess locomotive inconsistency in CNS rodents, for example, whenever the animal moves within a transparent plastic tube, its preliminary activity is registered as it rises against the stadium wall.
In the given question both technicians are correct because both are reliable ways to check cylinders and the influence of the belief if every pathway has many more advantages than each other.
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;
![p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905](https://tex.z-dn.net/?f=p%28t%3E5000%29%20%3D%20%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5000%7D%20%7B%5Clambda%20e%5E%7B-%5Clambda%20t%7D%7D%20%5C%2C%20dt%20%3D%5Cleft%20%5B%20%20-e%5E%7B%5Clambda%20t%7D%5Cright%20%5D_%7B5000%7D%5E%7B%5Cinfty%7D%20%3D%20e%5E%7B5000%20%5Clambda%7D%20%3D%200.5905)
(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) 
.
The difference between the electrical gradient is a electrical issue in power
Answer:
Continuous insulation helps in eliminating the necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.
Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.
Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.
Explanation:
Solution:
The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.
Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.
By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.
Continuous insulation helps building to withstand the test of time without the need to upgrade and repair
Now inside the placement of continuous insulation in the envelope is needed as follows:
- Foundation wall or slab insulation
- Insulation against sub floor
- Balcony interfaces
- On the interior of masonry wall
- Bond joist insulation
