Answer:
![\pi = \frac{A}{2r^2 + 2rh}](https://tex.z-dn.net/?f=%5Cpi%20%3D%20%5Cfrac%7BA%7D%7B2r%5E2%20%2B%202rh%7D)
Step-by-step explanation:
First, we need to isolate
by taking it common from both terms on the right:
![A=2\pi r^2 + 2\pi rh\\A=\pi (2r^2 + 2rh)](https://tex.z-dn.net/?f=A%3D2%5Cpi%20r%5E2%20%2B%202%5Cpi%20rh%5C%5CA%3D%5Cpi%20%282r%5E2%20%2B%202rh%29)
Now, since we want
in terms of the other variables, we can divide the left hand side (A) by whatever is multiplied with
on the right hand side. Then we will have an expression for
. Shown below:
![A=\pi (2r^2 + 2rh)\\\pi = \frac{A}{2r^2 + 2rh}](https://tex.z-dn.net/?f=A%3D%5Cpi%20%282r%5E2%20%2B%202rh%29%5C%5C%5Cpi%20%3D%20%5Cfrac%7BA%7D%7B2r%5E2%20%2B%202rh%7D)
This is the xpression for ![\pi](https://tex.z-dn.net/?f=%5Cpi)
Answer:
0.625
Step-by-step explanation:
Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.
Since A1 and A2 are mutually exclusive and exhaustive, we can say
b) P(B) = P(A1B)+P(A2B)
Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads.![P(B/A1) = 0.50 \\P(B/A2) = 0.75\\P(A1B) = 0.5(0.5) = 0.25 \\P(A2B) = 0.75(0.5) = 0.375\\P(B) = 0.625](https://tex.z-dn.net/?f=P%28B%2FA1%29%20%3D%200.50%20%5C%5CP%28B%2FA2%29%20%3D%200.75%5C%5CP%28A1B%29%20%3D%200.5%280.5%29%20%3D%200.25%20%5C%5CP%28A2B%29%20%3D%200.75%280.5%29%20%3D%200.375%5C%5CP%28B%29%20%3D%200.625)
c) Using Bayes theorem
conditional probability that it wasthe biased coin
=![\frac{0.375}{0.625} \\=\frac{3}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.375%7D%7B0.625%7D%20%5C%5C%3D%5Cfrac%7B3%7D%7B5%7D)
d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin=![\frac{0.25*0.5}{0.25*0.5+0.5*0.5} \\=\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%2A0.5%7D%7B0.25%2A0.5%2B0.5%2A0.5%7D%20%5C%5C%3D%5Cfrac%7B1%7D%7B3%7D)
Answer:
x < -4
Step-by-step explanation:
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