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2 days ago

Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba

bility that any particular couple or individual arrives late is 0.43 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let X = the number of people who arrive late for the seminar.
(a) Determine the probability mass function of X. [Hint: label the three couples #1, #2, and #3 and the two individuals #4 and #5.] (Round your answers to four decimal places.)
x P(X = x)
0
1
2
3
4
5
6
7
8

(b) Obtain the cumulative distribution function of X. (Round your answers to four decimal places.)
x F(x)
0
1
2
3
4
5
6
7
8

Physics

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Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

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We know that

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Now applying the conservation of momentum:

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$M = 0.400 kg$

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$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

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$x_1 = 0.01962 m$

$(0.300 + 0.400)9.81 = 200 x_2$

We also determine that the speed of SHM is represented as

$x_2 = 0.0343 m$

Here, we derive values from

$v = \omega\sqrt{A^2 - x^2}$

$\omega = \sqrt{\frac{k}{m + M}}$

$\omega = \sqrt{\frac{200}{0.300 + 0.400}}$

$\omega = 16.9 rad/s$

$a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m$

Using the previous formula gives:

$0.95 = 16.9\sqrt{A^2 - 0.0147^2}$

$A = 0.0581 m$

Part b)

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$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{16.9}$

$T = 0.37 s$

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