I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.
A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.
Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.
The number of ways of drawing 2 aces and 3 2s is
and the number of ways of drawing 3 aces and 2 2s is the same,
so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.
Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is
The total number of hands that can be drawn is
Then the probability of getting a full house is
Step-by-step explanation:
When f(x) = f(x - 4), 3x² = 3(x - 4)².
=> x² = (x - 4)², x = -(x - 4), 2x = 4, x = 2.
The x-value is 2.
9514 1404 393
Answer:
(c) f(x) is an even degree polynomial with a positive leading coefficient.
Step-by-step explanation:
The leading terms of the two functions are ...
f(x): x² (even degree, positive coefficient: 1)
g(x): x³ (odd degree, positive coefficient: 1)
Then it is true that ...
f(x) is an even degree polynomial with a positive leading coefficient
9514 1404 393
Answer:
14 units
Step-by-step explanation:
The angle bisector divides the sides proportionally, so you have ...
(x+4)/8 = (2x+1)/12
3(x +4) = 2(2x +1) . . . . . . multiply by 24
3x +12 = 4x +2 . . . . . . . . eliminate parentheses
10 = x . . . . . . . . . . . subtract (3x+2)
Then BD = x+4 = 10 +4.
The length of BD is 14 units.
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<em>Additional comment</em>
The "triangle" cannot exist, as the side lengths are shown as 8, 12, and 35. The long side is too long. Nice math; bad geometry.
