Answer:
your answer is correct
Explanation:
You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.
Answer:
Rock and orange juice
Explanation:
The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.
An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.
<h3>
What is uphill slope?</h3>
Driving uphill suggests climbing a “positive” six percent slope . Driving downhill, the “rise” exists as a drop, so there is a “negative,” or downhill, slope (Figure B). When dealing with slope, a positive slope simply indicates uphill and a negative slope indicates downhill. An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.
If something or someone lives moving downhill or is downhill, they exist moving down a slope or are located toward the bottom of a hill. He headed downhill toward the river. adverb. If you communicate that something exists going downhill, you mean that it is becoming worse or less prosperous.
To learn more about uphill slope refer to:
brainly.com/question/13361896
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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;
![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) 
.
