Answer:
Step-by-step explanation:
Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.
The tangent passing through (6,2) would have equaiton of the form
for suitable m.
Because this line is tangent, distance of centre of circle namely origin is the radius 1.
i.e. 
the coordinates are the intersection of the tangent with the circle
They are (-0.162, 0.987) and (0.462,-0.887)
Answer:
<h2>
64/27</h2>
Step-by-step explanation:
If x and y be real numbers satisfying 2/x=y/3=x/y, then any two of the equation are equated as shown;
2/x = y/3 ... 1 and;
y/3 = x/y... 2
From equation 1, 2y = 3x ... 3
and from equation 2; y² = 3x ... 4
Equating the left hand side of equation 3 and 4 since their right hand sides are equal, we will have;
2y = y²
2 = y
y = 2
Substituting y = 2 into equation 3 to get the value of x;
2y = 3x
2(2) = 3x
4 = 3x
x = 4/3
The value of x³ will be expressed as (4/3)³ = 4*4*4/3*3*3 = 64/27
Answer: E(Y) = 1.6 and Var(Y)=1.12
Step-by-step explanation:
Since we have given that
X 0 1 2
P(X) 0.4 0.4 0.2
Here, number of games = 2
So, 
Since 
are independent variables.
so, ![E[Y]=2E[X]\\\\Var[Y]=2Var[X]](https://tex.z-dn.net/?f=E%5BY%5D%3D2E%5BX%5D%5C%5C%5C%5CVar%5BY%5D%3D2Var%5BX%5D)
So, we get that
![E(X)=0.4\times 0+0.4\times 1+0.2\times 2=0.8\\\\and Var[x]=E[x^2]-(E[x])^2\\\\E[x^2]=0\times 0.4+1\times 0.4+4\times 0.2=1.2\\\\So, Var[x]=1.2-(0.8)^2\\\\Var[x]=1.2-0.64=0.56](https://tex.z-dn.net/?f=E%28X%29%3D0.4%5Ctimes%200%2B0.4%5Ctimes%201%2B0.2%5Ctimes%202%3D0.8%5C%5C%5C%5Cand%20Var%5Bx%5D%3DE%5Bx%5E2%5D-%28E%5Bx%5D%29%5E2%5C%5C%5C%5CE%5Bx%5E2%5D%3D0%5Ctimes%200.4%2B1%5Ctimes%200.4%2B4%5Ctimes%200.2%3D1.2%5C%5C%5C%5CSo%2C%20Var%5Bx%5D%3D1.2-%280.8%29%5E2%5C%5C%5C%5CVar%5Bx%5D%3D1.2-0.64%3D0.56)
So, E[y]=2×0.8=1.6
and Var[y]=2×0.56=1.12
Hence, E(Y) = 1.6 and Var(Y)=1.12
