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
In the given you wrote -2, whereas it should be -2i
z = a + bi and in trigo form | z |.(cos Ф + i.sin Ф).
z = 0 -2i → | z |.(cos Ф + i.sin Ф)
Now let's calculate z:
z² = a²+b² → z² = 0² (-2.i)² → z = -2(i)² →z= -2(-1) → z = |2|
tan Ф = b/a = -2/0 → tan Ф → - ∞ ↔ Ф = -90° or Ф = 270°
then the - 2 I ↔ |2|(cos 270° + i.sin 270°)
Answer:
<u>240cm³</u>
Step-by-step explanation:
To find our answer we will <u>multiply</u> the <em>volume, width, and height</em>.
4 · 20 · 3 = 240. Therefore, <u>240cm³ is our answer.</u>
<em>Have a good day!</em>
Answer:
=50
Step-by-step explanation:
Answer:
1. 14 = 2 + 12
2. 16 = 4 + 12
3. 18 = 6 + 12
4. 20 = 8 + 12
Step-by-step explanation:
Super easy. All you do is replace the numbers in your table with the corresponding letter. In this case we have a table of s and f.
Example for row two: f = s + 12. Replace s with 4 ( 4 is from your s column so you would replace it with that) then solve and plug in your answer (When you solve your answer, it will go under f column).16 = 4 + 12 . f = 16, s = 4.
Formula = f = s + 12.
1. 14 = 2 + 12
2. 16 = 4 + 12
3. 18 = 6 + 12
4. 20 = 8 + 12
