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Bumek [7]
3 years ago
14

The answer to 1/4 times 4 in fluid ounces

Mathematics
2 answers:
mixer [17]3 years ago
6 0
1 gallon. Add the 1/4 up for each day and you have the fraction being 4/4 which translates to 1 because its whole. So it is one gallon
Kitty [74]3 years ago
3 0
The answer is 1



Hope this helped :)

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Step-by-step explanation:

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the probability density for a particle in a box is an oscillatory function even for very large energies. Explain how the classic
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Answer:

This is achieved for the specific case when high quantum number with low resolution is present.

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In Quantum Mechanics, the probability density defines the region in which the  likelihood of finding the particle is most.

Now for the particle in the box, the probability density is also dependent on resolution as well so for large quantum number with small resolution, the oscillations will be densely packed and thus indicating in the formation of a constant probability density throughout similar to that of classical approach.

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Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom
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Step-by-step explanation:

Given that:

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(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))

= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}

=  1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

=  1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}

Now P(X(s) \leq 0) = P(X(s) = 0)

(b)  We can Determine the conditional mean E[W3|X(t)=5] as follows;

E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\  = 1- P (X(s) \leq 2 | X (t) = 5 )  \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}

Now; P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of W_2 given that  X(t)=5 is:

f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 )  \\ \\  = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)

7 0
3 years ago
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:0

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Answer:

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Step-by-step explanation:

the standard form of a line is is usually given as Ax + By = C, so just by deduction you can tell that the correct answer is a.

If you want to do the procedure it would be:

y − 3 = 1/2 (x + 6)

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y - 1/2 x = 6

(y - 1/2 x = 6) * 2 to get rid of the fraction

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x - 2y = -12

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3 years ago
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