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The load that the same type of beam can support is 583.345 pounds if its width is 3.5 inches.
As the maximum load that can be carried by the beam is directly proportional to the width of the beam and a 1.5 inches wide beam can support a load of 250 pounds, an equation can be written as;
L = kw
Here L represents the maximum load supported and w represents the width and k is the constant of proportionality. Therefore;
250 = k × 1.5
Solving for k;
k = 250 ÷ 1.5
k = 166.67
Therefore the equation can be written as;
L = 166.67 w
As the width of the same type of beam is 3.5 inches, the maximum load it can support can be calculated as follows;
L = 166.67 × 3.5
L = 583.345 pounds
Therefore; 583.345 pounds is the maximum load that the beam can support if its width is 3.5 inches.
To learn more about the constant of proportionality, click here:
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Many of the substances in welding smoke, such as chromium, nickel, arsenic, asbestos, manganese, silica, beryllium, cadmium, nitrogen oxides, phosgene, acrolein, fluorine compounds, carbon monoxide, cobalt, copper, lead, ozone, selenium, and zinc, can be extremely toxic.
Answer:
0.14% probability of observing more than 4 errors in the carpet
Explanation:
When we only have the mean, we use the Poisson distribution.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.
This means that ![\mu = 0.8](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.8)
What is the probability of observing more than 4 errors in the carpet
Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So
![P(X \leq 4) + P(X > 4) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%2B%20P%28X%20%3E%204%29%20%3D%201)
We want P(X > 4). Then
![P(X > 4) = 1 - P(X \leq 4)](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29)
In which
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.4493)
![P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.3595)
![P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.1438)
![P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.0383)
![P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8%7D%2A%280.8%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.0077)
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.4493%20%2B%200.3595%20%2B%200.1438%20%2B%200.0383%20%2B%200.0077%20%3D%200.9986)
![P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29%20%3D%201%20-%200.9986%20%3D%200.0014)
0.14% probability of observing more than 4 errors in the carpet