$\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}$

$Answer:\ \dfrac{19}{52}$

5 0

3 years ago

A choreographer uses a number line to position dancers for a ballet. Dancers A and B

kenny6666 [7]

First we will calculate the distance AB.
$AB=B-A=23-5=18.$
If C is 5/9 of the distance, mathematically speaking that means:
$C= \frac{5}{6} *AB= \frac{5}{6} *18=15.$
R: The coordonate of C is 15.

5 0

2 years ago

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