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Svetlanka [38]
3 years ago
14

A six-sided number cube labeled from 1 to 6 is rolled. What is the probability of getting a multiple of 2 or a multiple of 3?

Mathematics
2 answers:
Hoochie [10]3 years ago
8 0

Answer:

4/6 I think

Step-by-step explanation:


luda_lava [24]3 years ago
5 0

Answer:

2/3

Step-by-step explanation:

A six-sided number cube labeled from 1 to 6 is rolled.

Total outcomes = 1,2,3,4,5,6= 6 outcomes

multiple of 2  are 2,4,6 so 3 outcomes

multiple of 3 are 3,6 so 2 outcomes

multiply of 2 or 3  are 2,4,3,6 so 4 outcomes

the probability of getting a multiple of 2 or a multiple of 3

= possible outcomes / total outcomes

= 4/6

now reduce the fraction, divide by 2 on both sides

the probability of getting a multiple of 2 or a multiple of 3 = 2/3

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a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

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Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

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