I believe they are similar and the ration would be 2:1, correct me if I'm wrong
Modelling can be done by Graph Data Modelling
Step-by-step explanation:
1.Graph data modeling is the process in which a user describes an arbitrary domain as a connected graph of nodes and relationships with properties and labels.
2.The following tasks are performed in an iterative manner:
Identify entity types.
Identify attributes.
Apply naming conventions.
Identify relationships.
Apply data model patterns.
Assign keys.
Normalize to reduce data redundancy.
Denormalize to improve performance.
3.One such approach is the property graph model, where data is organized as nodes, relationships, and properties (data stored on the nodes or relationships). ... Nodes are the entities in the graph. They can hold any number of attributes (key-value pairs) called properties
Answer:
44%
Why? Because there is only 4 of each and you have so many more chances to pull a different card.
Very nice to have an accompanied image!Illumination is proportional to the intensity of the source, inversely proportional to the distance squared, and to the sine of angle alpha.so that we can writeI(h)=K*sin(alpha)/s^2 ................(0)where K is a constant proportional to the light source, and a function of other factors.
Also, radius of the table is 4'/2=2', therefore, using Pythagoras theorem,s^2=h^2+2^2 ...........(1), and consequently,sin(alpha)=h/s=h/sqrt(h^2+2^2)..............(2)
Substitute (1) and (2) in (0), we can writeI(h)=K*(h/sqrt(h^2+4))/(h^2+4)=Kh/(h^2+4)^(3/2)
To get a maximum value of I, we equate the derivative of I (wrt alpha) to 0, orI'(h)=0or, after a few algebraic manipulations, I'(h)=K/(h^2+4)^(3/2)-(3*h^2*K)/(h^2+4)^(5/2)=K*sqrt(h^2+4)(2h^2-4)/(h^2+4)^3We see that I'(h)=0 if 2h^2-4=0, giving h=sqrt(4/2)=sqrt(2) feet above the table.
We know that I(h) is a minimum if h=0 (flat on the table) or h=infinity (very, very far away), so instinctively h=sqrt(2) must be a maximum.Mathematically, we can derive I'(h) to get I"(h) and check that I"(sqrt(2)) is negative (for a maximum). If you wish, you could go ahead and find that I"(h)=(sqrt(h^2+4)*(6*h^3-36*h))/(h^2+4)^4, and find that the numerator equals -83.1K which is negative (denominator is always positive).
An alternative to showing that it is a maximum is to check the value of I(h) in the vicinity of h=sqrt(2), say I(sqrt(2) +/- 0.01)we findI(sqrt(2)-0.01)=0.0962218KI(sqrt(2)) =0.0962250K (maximum)I(sqrt(2)+0.01)=0.0962218KIt is not mathematically rigorous, but it is reassuring, without all the tedious work.