Answer:
If it is less than 3, Player 1 earns 3 points.
If not, Player 2 earns 2 points.
Step-by-step explanation:
<u>Player 1</u> :
p(N < 3) = p(N = 1 or N = 2) = 2/5
<u>Player 2</u> :
p(N ≥ 3) = p(N = 3 or N = 4 or N = 5) = 3/5
<u>We notice that</u> :
p(N < 3) × 3 = (2/5) × 3 = 6/5
On the other hand,
p(N ≥ 3) × 2 = (3/5) × 2 = 6/5
since ,the probability player 1 win multiplied by the associated number of points (3)
is equal to
the probability player 2 win multiplied by the associated number of points (2).
Then the game is fair.
Yes, you can; based on the inherent assumption that the "two radicals that have negative values" are, in fact, "imaginary numbers" .
Take, for example, the commonly known "imaginary number": "i" ; which represents the "imaginary number" ; " √-1 " .
Since: "i = √-1" ;
Note that: " i² = (√-1)² = √-1 * √-1 = √(-1*-1) = √1 = 1 .
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10 + (-5) = 10 - 5 = 5
-11 + 20 = 20 + (-11) = 20 - 11 = 9
102 + (-1) = 102 - 1 = 101
<span>-20 + 60 = 60 + (-20) = 60 - 20 = 40</span>
Answer:
The factor of 12x^2 + 16 is 4x (3x+4)
