Explanation:
Given T = 10 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (10 + 273.15) K = 283.15 K
<u>T = 283.15 K </u>
The conversion of T( °C) to T(F) is shown below:
T (°F) = (T (°C) × 9/5) + 32
So,
T (°F) = (10 × 9/5) + 32 = 50 °F
<u>T = 50 °F</u>
The conversion of T( °C) to T(R) is shown below:
T (R) = (T (°C) × 9/5) + 491.67
So,
T (R) = (10 × 9/5) + 491.67 = 509.67 R
<u>T = 509.67 R</u>
Answer:
The theoretical maximum specific gravity at 6.5% binder content is 2.44.
Explanation:
Given the specific gravity at 5.0 % binder content 2.495
Therefore
95 % mix + 5 % binder gives S.G. = 2.495
Where the binder is S.G. = 1, Therefore
Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495
(95 +5)/(Vx +5) = 2.495
2.495 × (Vx + 5) = 100
Vx =35.08 to 95
Or density of mix = Mx/Vx = 95/35.08 = 2.7081
Therefore when we have 6.5 % binder content, we get
Per 100 mass unit
93.5 Mass unit of Mx has a volume of
Mass/Density = 93.5/2.7081 = 34.526 volume units
Therefore we have
At 6.5 % binder content.
(100 mass unit)/(34.526 + 6.5) = 2.44
The theoretical maximum specific gravity at 6.5% binder content = 2.44.
Answer:
- Depletes Fossil Fuels
- Polution
Explanation:
Fossil Fuels take millions of years to be made. Gasoline is made up of primarily petroleum (a fossil fuel) which is easily refined into gasoline or kerosene. This means that once you run out of a fossil fuel...you're out of it.
Gas cars are also very big polutants. If you've ever heard of an emmisions test for a car, it is a test that sees how much pollution a car puts out into the air. It meassures how safe those levels are. This is a big indicator of how much cars pollute becasue there is a need to test your car every year. The gases that can be put into the air are harmful not only to humans and animals but to the climate and atmosphere.
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) 
.
