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natta225 [31]
2 years ago
14

2) What kinds of food can you eat in space?

Engineering
2 answers:
shtirl [24]2 years ago
6 0

Answer:

fruits, nuts, peanut butter, chicken, beef, seafood, candy, brownies now  if we talking about drinks ,coffee, tea, orange juice, fruit punches and lemonade

Explanation:somehow this all are delicious

rjkz [21]2 years ago
5 0

Hi there, so what I know is:

An astronaut can choose from many types of foods such as fruits, nuts, peanut butter, chicken, beef, seafood, candy, brownies, etc. Available drinks include coffee, tea, orange juice, fruit punches and lemonade. As on Earth, space food comes in disposable packages.

--------------------

<em>I hope this helps, and I apologize if I get the answer wrong for you, I am 100 percent hopeful that my answer is correct. However if it isn't, please tell me and I will make up for my mistakes. Thank you and have a great rest of your day or night! :></em>

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Technician A says that the first step in diagnosing engine condition is to perform a thorough visual inspection. Technician B sa
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2 years ago
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have
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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) = \lambda e^{-\lambda t}

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1

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

(c) The Poisson distribution is presented as follows;

P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!}  = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!}  = 3.915\times 10^{-12}

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) CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}.

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