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aleksandrvk [35]
3 years ago
15

Of the 52 cards in the deck, four are aces and others (kings, queens, jacks, and tens) are worth 10 points each. The dealer has

a blackjack if one card is an ace and the other is worth 10 points; it doesn't matter which card is face up and which card is face down. How many different blackjack hands are there?
Mathematics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

64 different black jack hands

Step-by-step explanation:

there are 4 aces in a deck, and you can score a blackjack if any of these 4 aces is paired with any of the 16 different cards that are worth 10 points. Total number of different blackjack combinations = 4 aces x 16 different tens = 64.

Even though the number of possible blackjack hands might seem large, the possibilities of actually getting one are really low:

possibility of getting 1 ace = 4/52 = 1/13

possibility of getting 1 ten = 16/51

possibility of getting a blackjack = 16 / 663

if you start with the 10 first the odds are the same:

possibility of getting 1 ten = 16/52 = 4/13

possibility of getting 1 ace = 4/51

possibility of getting a blackjack = 16 / 663

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-2x^2 + 8x - 16 + 32/x+2
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What is the interquartile range of the data set? {18, 17, 10, 25, 20, 20, 30, 18, 18, 30, 25} Enter the answer in the box. PLEAS
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The interquartile is 21
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PLEASE HELP ASAP!!!!
musickatia [10]
Let us model this problem with a polynomial function.

Let x =  day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.

Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅

Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
|   1     1    1    1    1  |  | a₁ |      | 42 |
| 2⁴  2³  2²  2¹  2⁰  |  | a₂ |      | 26 |
| 3⁴  3³  3²  3¹  3⁰   |  | a₃ |  =  | 61 |
| 4⁴  4³  4²  4¹  4⁰   |  | a₄ |      | 65 |
| 5⁴  5³  5²  5¹  5⁰  |  | a₅ |      | 56 |

When this matrix equation is solved in the calculator, we obtain
a₁ =      4.1667
a₂ =  -55.3333
a₃ =  253.3333
a₄ =  -451.1667
a₅ =   291.0000

Test the solution.
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y(3) = 61
y(4) = 65
y(5) = 56

The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.

Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631


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The range = highest - lowest value

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= 41.

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