A hostile person may have poorer health outcomes because:

A personality characterized by anxiety, depression, and lack of emotional expression has been called:

In a series of studies described in lecture, the participants that ate the most ice cream were:

In the North article, “Could a Personality Test Improve Your Health?”, the author provides support for the idea that understanding and possibly changing one’s personality could:

In a study by Mann and her grad student looking at HPA activation and dieting, they found:

  Based on what you have learned in this class, which of the following are likely to influence your goal to eat more fruits and vegetables:    

According to the Agency for Healthcare Research & Quality, a Patient-centered Medical Home (PCMH) provides care that is:

Research on mHealth technologies has found:

In studies of personality and health:

You are given a collection of directed, weighted edges that represent a graph. Each edge connects two nodes and has a weight associated with it. An edge is defined using the following class: public class Edge {    public E src;    public E dest;    public int weight;} Implement a method with the signature: public Queue findGreedyPath(List edgeList, E src, E dest) that does the following: Given src and dest nodes, return a Queue of Edges that represents a path from src to dest in which each edge is the lightest (lowest weight) edge coming out of each node in the path that does not result in a cycle, i.e. does not go back to a node that is already in the path. For instance, let’s say you have the following graph and have “A” as the src and “E” as the dest:     The method should work as follows: We start at the src, which is A. The lightest edge coming out of A is the one to C, which has a weight of 4, so we would add that Edge to the Queue; note that although there is an edge to our dest node, we don’t select that one since it’s not the lightest Now we’re at C. There is only one edge coming out of it, which goes to D, so we add that Edge to the Queue. Now we’re at D. The lightest edge is the one going to A, but we have already visited A, so we look for the next lightest edge, which is the one going to E, and we add that Edge to the Queue. We’re at E, which is the dest, so we’re done, we return the Queue containing the three Edge objects. You may assume that: the graph is completely connected, i.e. every node is reachable from every other node, which means that every node has at least one Edge. edge weights are always positive all edges coming out of a node have distinct weights, i.e. there will never be a tie for the lightest edge you do not need to check for null inputs values or any other error conditions in your implementation Last, you may also assume that a solution always exists for every src-dest pair in the graph; you do not need to worry about the case when all of a node’s outgoing edges go to nodes that are already in the path. This question will be graded by a member of the instruction staff; partial credit is possible. It is okay if there are slight compilation errors in your code, but you will lose points for using the wrong methods Java classes that you use. You may consult https://docs.oracle.com/javase/8/docs/api to help find the right methods.