Answer:
Technician A
Explanation:
1- It's was easy to understand.
2- Holding down a battery could actually cause battery plate damage. Most of the batteries I interact with only use a plastic cover and a small spring to hold the battery in place.
-Hope this helped :) -
Answer:
Explanation:
(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄
Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14
1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54
= 153.12 + 21.28 + 148.16 + 7.56
= 330.12 g/mol
which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable
The moles of methane produced will be given as
m = (4a + b -2c - 3d)/8
= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8
= (51.04 + 21.28 - 18.52 - 1.62)/8
= 52.18/8
= 6.5225
(b) Volume of methane V is given as
V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)
= 0.1461 m³ CH₄/kg lawn trimmings
(c) Energy will be given as
CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol
= 5805.025
≈ 5805 kJ/kg
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) 
.
Answer: Yes
Explanation:
While handling and managing offices, means a lot of different tasks, including files,overseeing supplies, running the mailroom and worker productivity, the space planner work is to specifically focus on managing the space within your office building. When it comes to all room within each wing of each floor, the space planner's work is to layout and optimize the design in order to maximize productivity and also keep the employees happy.
Work of the space ePlanner includes all those in listed above in the question.
