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3 years ago

A proposed design for a habitable Mars colony is a hemispherical

Mathematics

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

r = 1.37 billion feet

Step-by-step explanation:

It is given that,

A proposed design for a habitable Mars colony is a hemispherical biodome used to maintain a breathable atmosphere for the colonists.

The volume of biodome is 5.5 billion cubic feet of air.

It is in the shape of a hemisphere. The volume of a hemisphere is given by :

$V=\dfrac{2}{3}\pi r^3$ , r = radius

V = 5.5 billion cubic feet

$r^3=\dfrac{3V}{2\pi}\\\\r^3=\dfrac{3\times 5.5}{2\times 3.14}\\\\r=(\dfrac{3\times 5.5}{2\times 3.14})^{1/3}\\\\r=1.37\ \text{billion feet}$

So, the radius of the biodome is 1.37 billion feet.

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Answer:

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Step-by-step explanation:

From the given information:

Assume F is used to denote the two cards;

If there are four aces among 52 playing cards, the chance of selecting the first ace is = $\dfrac{4}{52}$

After selecting the first ace, we have only 3 aces remaining and a total of 51 playing cards. Thus, the chance of selecting the second ace will be $\dfrac{3}{51}$

By applying product rule, we can determine the chance of selecting two aces without replacement as follows:

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$P(F=22)=\dfrac{4}{52}\times \dfrac{3}{51}$

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The probability of getting one ace, one face is:

$P(F=21) =( \dfrac{4}{52}\times \dfrac{12}{51})+( \dfrac{12}{52}\times \dfrac{4}{51})$

$P(F=21) = \dfrac{4}{221}\times\dfrac{4}{221}$

$P(F=21) = \dfrac{8}{221}$

Since there are 4 aces, 4 nine, and 12 faces in a card deck

The probability of getting one ace, one nine, or two faces now will be:

$P(F=20) = (\dfrac{4}{52} \times \dfrac{4}{51})+ (\dfrac{4}{52} \times \dfrac{4}{51}) + (\dfrac{12}{52} \times \dfrac{11}{51})$

$P(F=20) = (\dfrac{53}{663} )$

Now, the probability of at least 20 now is:

$\text{P(F at least 20)} = \dfrac{1}{221}+\dfrac{8}{221}+\dfrac{53}{663}$

$\text{P(F at least 20)} = \dfrac{80}{663}$

If H represents the amount of prize of the expected winnings:

Then;

$(H - 10) (\dfrac{80}{663}) + (-10)(\dfrac{663-80}{663}) = 0$

$\dfrac{80(H-10)}{663}-\dfrac{5830}{663}=0$

$\dfrac{80(H-10)}{663}=\dfrac{5830}{663}$

80H - 800 = 5830

80H = 5830 +800

80H = 6630

H = 6630/80

H = $82.875

The prize should be $82.875 to make a winning positive.

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