ANSWER

EXPLANATION
Lines WO and WX are tangents to the circle.
The angle subtended at the center, Z by minor arc XO is the angle subtended at the circumcenter by the same arc.
Therefore the central angle is 2x°.
The two radii ZX and ZO meets the tangents at right angles.
This implies that:

Or



Divide both sides by 2

I believe so, not 100% sure tho
If the original population has a Normal distribution, then the distribution of the sample mean is also Normal.
If we extract random samples of size n from a population with half a μ and standard deviation σ, then the distribution of the sample mean will have the following parameters:
μ_Xbar = μ
σ_Xbar = σ/ √(n).
Where:
μ = 50
σ = 4
n = 20
So:
μ_Xbar = 50
σ_Xbar = 0.8944
See the attached figure to better understand the problem
we know that
AM=MD
so
triangle AMD is an isosceles right triangle
therefore
its height is half its width.
Then
AB = (1/2)AD----------Equation 1
Perimeter=2*[AB+AD]=34 in ---------> AB+AD=17-------> Equation 2
I substitute 1 in 2
(1/2)AD +AD = 17
(3/2)AD=17
AD=17*2/3----------> AD=34/3-------> 11 1/3 in
AB=(1/2)AD--------> AB=(1/2)*34/3--------> AB=17/3-------> AB=5 2/3 in
the answers are
AD=11 1/3 inAB = 5 2/3 in
You would lay out the equation for area
A=(L)(W)
then plug in the variables
8x^2+26x+12=2x+5(L)
then solve for L