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

Answer:
• David
,
• 4 miles
Explanation:
In the graph:
The given locations are:
• Owen's House, A(11,3)
,
• David's House, B(15,13)
,
• School, C(3,18)
We determine both Owen's and David's distance from the school using the distance formula.

Owen's distance from school (AC)
![\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AC%3D%5Csqrt%5B%5D%7B%283-11%29%5E2%2B%2818-3%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-8%29%5E2%2B%2815%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B64%2B225%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B289%7D%20%5C%5C%20AC%3D17%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
David's distance from school (BC)
![\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20BC%3D%5Csqrt%5B%5D%7B%283-15%29%5E2%2B%2818-13%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-12%29%5E2%2B%285%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B144%2B25%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B169%7D%20%5C%5C%20BC%3D13%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
We see from the calculations that David lives closer to the school, and by 4 miles.
The graph below is attached for further understanding:
Answer:
angle DFG = 49 degrees
angle JKL = 41 degrees
Step-by-step explanation:
When angles are complementary with each other, it means that if you add both of the angles up, it adds up to 90 degrees.
In this question, you would have to add up angle DFG and angle JKL and find the x that makes the equation equal to 90 degrees.
angle DFG = x + 5
angle JKL = x - 3
(x + 5) + (x - 3) = 90
2x + 2 = 90
2x = 90 - 2
2x = 88
x = 44
But since we have to find out the angle measures, we have to the "x = 44" with the x's in the DFG and JKL angles.
DFG = (44) + 5 = 49
JKL = (44) - 3 = 41
Assuming that the label only covers the body of the cylinder and not the circular faces on the top and bottom.