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I am Lyosha [343]
3 years ago
14

Given P(X) = 0.5, P(Y) = 0.4, and P(Y|X) = 0.3, what are P(X and Y) and P(X or Y)?

Mathematics
1 answer:
scZoUnD [109]3 years ago
6 0

Use the definition of conditional probability to find the "and" probability:

P(Y\mid X)=\dfrac{P(X\cap Y)}{P(X)}\implies P(X\cap Y)=0.3\times0.5=0.15

Use the inclusion/exclusion principle to find the "or" probability:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)=0.5+0.4-0.15=0.75

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A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4
vlabodo [156]

Answer:

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

Step-by-step explanation:

Bernoulli's equation,

P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2

P₁ = P₂= atmospheric presser

\rho= density

\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2             [since P₁ = P₂]

\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)

\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2

\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2

\Rightarrow v^2_2- v^2_1=2g h                                [h_1-h_2=h]

Here   v_1\approx 0

\Rightarrow v^2_2=2g h

\therefore v_2=\sqrt {2gh

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole v_2 = \sqrt {2\times 9.8\times 10.4} m/s

                                                                                  =14.28 m/s.

Given, the rate of flow from the leak is 2.53\times 10^{-3} m^3/min

                                                               =\frac{2.53\times 10^{-3}}{60}  m^3/s

Let the diameter of the hole be d.

Then the cross section area of the hole is =\pi (\frac d2)^2

We know that,

The rate of flow = Cross section area × speed

\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28

\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}

\Rightarrow d= 1.94 \times 10^{-3}

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

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