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Neko [114]
4 years ago
13

If a coin is tossed 3 times, and then a standard six-sided die is rolled 4 times, and finally a group of four cards are drawn fr

om a standard deck of 52 cards without replacement, how many different outcomes are possible?
Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
3 0

Answer:

67,365,043,200

Step-by-step explanation:

A coin toss has 2 possible outcomes.  A coin tossed 3 times has 2³ = 8 possible permutations.

A standard die has 6 possible outcomes.  A die rolled 4 times has 6⁴ = 1296 possible permutations.

The number of ways 4 cards can be chosen from a deck of 52 without replacements is 52×51×50×49 = 6,497,400.

The total number of possible outcomes is:

8 × 1296 × 6,497,400 = 67,365,043,200

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Let´s call   x  and y the sides of the rectangular area

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C = Costs ( sides x)  + costs ( sides y )

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Cost of sides y    =  10*y  +  10 *y   =  2*10*y  = 20*y

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The function cost as function of x is:

C(x)  = 22*x + 20*(990/y)  =  22*x  +  19800/x

Tacking derivative on both sides of the equation:

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Solving for x

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22x²  =  19800

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x₁,₂ = ± 30        We dismiss negative root ( we never have negative lenghts)

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C´´ (x) = 39600/x³    so C´´ is always greater than 0 then C (x) has a minimum at x = 30

Sides of the rectangular area are:

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y  =  33

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