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
<span>1. </span><span>Given equation = 2 thousands 7
tens / 10.
From this given equation, give the unit form and the standard form
Unit form is the expression where you replace the number with its place values.
=> 2 thousands and 7 tens divided by 1 tens.
=> 2 hundreds 7 ones – answer in unit form
Standard form is the expression which uses numbers and operators to express an
equation.
=> 2 070 / 10
=> 207 – answer in standard form</span>
Answer:
145°
Step-by-step explanation:
There are a couple of ways you can get there:
1. ∠ACB is a right angle, 90°. Hence, ∠BAC is the complement of ∠ABC, so is ...
... ∠BAC = 90° -∠ABC = 90° -55° = 35°
Then, ∠BAC and ∠BAD are a linear pair, so total 180°. That makes ∠BAD the supplement of ∠BAC, so ...
... ∠BAD = 180° -35° = 145°
2. ∠BAD is the exterior angle at A for the triangle ABC. It will have a measure that is the sum of the opposite interior angles: given ∠ABC = 55° and right angle ACB = 90°.
... ∠BAD = 55° +90° = 145°
You could draw a 2•5 5•2 10•1 and 1•10
I think that would be a rhombus.
