If a= -3 what is 3a^2
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Let's call the aces for hearts, diamonds, clubs and spades. So,
are red and [ted] c, s[/tex] are black.
Since the first card is replaced, the two picks are identical. This means that the sample space is given by all the possible couple
There are 16 such couples (we have four choices for the first card, and the same four choices for the second card). Now let's compute the odds in our favour to deduce the probability of winning:
If we want a player to draw two card of the same colour, the following couples are good:
so 8 possible couples over 16. This means that the probability that a player draws two cards of the same color is 8/16 = 1/2.
Similarly, the probability of drawing a red ace first and then a black ace is represented by the following couples:
which are 4 over the same 16 as above, thus leading to a probability of 4/16 = 1/4.
Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)
The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)
At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots
The distance between the ships is increasing at about 28.37 knots at 4 pm.
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)
The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)
At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots
The distance between the ships is increasing at about 28.37 knots at 4 pm.