We know P = 1/2
X~B(2, 0.5)
n = 1
2C1 from pascal triangle = 2
P(X=0.5)= 2C1 x 0.5 x 0.5
= 0.5
= 50%
![\bf \begin{array}{clclll} -6&+&6\sqrt{3}\ i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A-6%26%2B%266%5Csqrt%7B3%7D%5C%20i%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0Aa%26%26b%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%0A%5Ctheta%20%3Dtan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bb%7D%7Ba%7D%20%5Cright%29%0A%5Cend%7Bcases%7D%5Cqquad%20r%5Bcos%28%5Ctheta%20%29%2Bi%5C%20sin%28%5Ctheta%20%29%5D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.
thus
Dividing the 2nd equation by 2 gives y=2x + 3 which is the same as the too
p line. any answer to one of them is obviously an answer to the other so there are infinitely many solutions
Answer:
<h2>3x - y + 7 = 0</h2>
Step-by-step explanation:
The slope-intercept form of an equation of a line:

m - slope
b - y-intercept
Put the given y-intercept b = 7 and the coordinates of the point (-2, 1) to the equation:
<em>subtract 7 from both sides</em>
<em>divide both sides by (-2)</em>

We have the equation:

Convert it to the general form
:
<em>subtract 3x and 7 from both sides</em>
<em>change the signs</em>

70 ft because 7 is bigger than 5 and 7x10=70