Optio D is the correct one. .pdf

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Optio D is the correct one. Solution Optio D is the correct one..

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The probability of atom has zero quanta of energy is P=Number.pdf

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Copper turns green because it oxidizes. Coppe.pdf

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CO2 and NO2 are acidic oxides as non-metal form acidic oxides. Al2O3 is an amphoteric oxide as it is a lewis acid but forms Al(OH)3 (which is basic) in water. CaO is basic oxide as metals form basic oxides Solution CO2 and NO2 are acidic oxides as non-metal form acidic oxides. Al2O3 is an amphoteric oxide as it is a lewis acid but forms Al(OH)3 (which is basic) in water. CaO is basic oxide as metals form basic oxides.

CO2 and NO2 are acidic oxides as non-metal form a.pdf

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Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. These are appeared in sea stars, these openings are smaller, and yellow button like warts located on the aboral surface of sea stars. This opening has other important parts, which perform meticulous function in water vascular system, and these parts are tube feet, stone canal, ring and radial canals. Ques-2: Spine structure of Ophiuroidea (echinoderm) are arranged in 5 rows i& connected to ligaments with mutable collagenous substance. These spines are composed of dermal calcareous ossicles or calcareous spicules Solution Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. These are appeared in sea stars, these openings are smaller, and yellow button like warts located on the aboral surface of sea stars. This opening has other important parts, which perform meticulous function in water vascular system, and these parts are tube feet, stone canal, ring and radial canals. Ques-2: Spi.

Ques-1 Development of tube feet of CRINOIDEADevelopment of tube f.pdf

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Program.csusing System; using System.Collections.Generic; usin.pdf

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package chegg; import java.util.ArrayList; import java.util.List; import java.util.ListIterator; import java.util.Scanner; import java.util.Set; import java.util.TreeMap; public class Vowles { public static void main(String args[]) { /* creating instance variable of List */ List letters = new ArrayList(); /* vowels instance variable for holding vowels */ List vowels = new ArrayList(); /* creating HashMap for storing vowels and count */ TreeMap map = new TreeMap(); /* Inserting vowels into treemap */ map.put(\"a\", 0); map.put(\"e\", 0); map.put(\"i\", 0); map.put(\"o\", 0); map.put(\"u\", 0); /* * creating instance variable of Scanner class for taking input from * console */ Scanner scanner = new Scanner(System.in); /* * prompt for user input */ System.out.println(\"Please enter the letters with comma seperated\"); String input = scanner.nextLine(); /* closing scanner object */ scanner.close(); String[] letter = input.split(\",\"); /* Inserting letters into List */ for (String values : letter) { letters.add(values); } /* ListIterator for iterating list */ ListIterator itr = letters.listIterator(); while (itr.hasNext()) { String val = itr.next(); /* checking if value is any vowels */ if (val.equals(\"a\") || val.equals(\"e\") || val.equals(\"i\") || val.equals(\"o\") || val.equals(\"u\")) { /* If vowels already exist in tree map */ if (map.containsKey(val)) { Integer value = map.get(val); map.put(val, ++value); vowels.add(val); } else { map.put(val, 1); vowels.add(val); } } } /* Removing all element from letters list */ letters.removeAll(letters); /* Getting size of vowels */ int listSize = vowels.size(); System.out.println(\"There are \" + listSize + \" vowels in the list:\"); /* Iterating treeMap */ Set keys = map.keySet(); for (String key : keys) { System.out.print(key + \"- \" + map.get(key) + \" \"); } } } /***************output******************/ Please enter the letters with comma seperated a,f,v,q,e,a,o,g,y,h,e,q,u There are 6 vowels in the list: a- 2 e- 2 i- 0 o- 1 u- 1 please rate the answer as best answer. Thanks Solution package chegg; import java.util.ArrayList; import java.util.List; import java.util.ListIterator; import java.util.Scanner; import java.util.Set; import java.util.TreeMap; public class Vowles { public static void main(String args[]) { /* creating instance variable of List */ List letters = new ArrayList(); /* vowels instance variable for holding vowels */ List vowels = new ArrayList(); /* creating HashMap for storing vowels and count */ TreeMap map = new TreeMap(); /* Inserting vowels into treemap */ map.put(\"a\", 0); map.put(\"e\", 0); map.put(\"i\", 0); map.put(\"o\", 0); map.put(\"u\", 0); /* * creating instance variable of Scanner class for taking input from * console */ Scanner scanner = new Scanner(System.in); /* * prompt for user input */ System.out.println(\"Please enter the letters with comma seperated\"); String input = scanner.nextLine(); /* closing scanner object */ scanner.close(); String[] letter = in.

package chegg;import java.util.ArrayList; import java.util.List;.pdf

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B. propene. Solution B. p.pdf

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$L{e^{-t}} = int _0 to infinty [e^{st}e^{2-t} dt = int _0 to infinty.pdf$ $L{e^{-t}} = int _0 to infinty [e^{st}e^{2-t} dt = int _0 to infinty.pdf$

L{e^{-t}} = int _0 to infinty [e^{st}e^{2-t} dt = int _0 to infinty.pdf

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Issue of Equity or Debt are commonly used methods for business financing. Buyer\'s of the Comany\'s Debt instruments have no claim to ownership in the business. This prevents dilution of the Company\'s ownership. The holder of the debt instrument is akin to a lender being entitled to principal / face value / base value and interest payments on the said amount at the specified rate. The holder of the debt instrument in the Company has no right to the future profits of the Company. However, due to the fixed nature of the returns on these instruments, risk borne by the holder is low. Hence, the Cost of Debt Capital is also lower than Equity Capital. Issue of New Equity dilutes the existing ownership interest of the Company. In most cases it also increases the shareholder base (number of shareholders) of the Company. The major disadvantage is the possibility of future loss of control and decision making power (if issued to large investors). However, the Company is not obligated to make payments of interest regularly as in the case of Debt Instruments. This in turn preserves the cash flow of the Company and makes profits available for reinvestment into the business. Equity holders take on more risk; due to the uncertainty of business success and growth leading to uncertainty of cash inflows. This is further magnified under a complex capital structure where the Company may have several other obligations to meet in the pecking order first, such as secured debt, convertible hybrid instruments, etc. The Company may also have obligations to Government Departments. These factors result in Equity holders demanding a higher rate of return, for absorbing this risk. Thus, the Cost of Equity is higher than cost of Debt. When new debt is issued at high finance costs, i.e. high rates of interest, the Company may face additional turmoil during uncertain or difficult financial situations as it increase the risk of insolvency. Where the Company is already highly leveraged (high debt to equity ratio) and new debt is issued at a high rate, the Company may face slower growth rates because of the obligation to service debt before pumping resources into high growth projects and endeavours. Further, the cost of debt also determines the cash flow to equity. However, the one of the biggest advantages of debt over equity is that interest cost is allowed as a deduction for the computation of tax. This in turn lowers the tax liability. Thus, the actual cost of debt is lower than stated cost by the amount of the tax saving benefit. The management of the Company is required to provide regular information for the investors / shareholders to monitor. Investors / Shareholders regularly seek comprehensive updates about the business, financial and performance information, guidace for the future, etc. This is especially true for newer investors that are still gaining comfort about the Company This is largely avoided when Debt instruments are issued. Raising equity finance is very d.

Issue of Equity or Debt are commonly used methods for business finan.pdf

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In probability we have two types of events: Independent: Events who\'s probability is not affected by the previous event(s). Dependent: Events who\'s probability can be affected by preious event(s). What we have in this case is a dependent event because when we don\'t replace the marble, it changes the amount of total marbles inside the jar and thus the chances of drawing another marble changes. Because one event relys on the other we must find their probabilities separately at first. So what we do is figure out the probability of pulling out a blue marble and then the probability of pulling out another after not replacing the first. We will call these event 1 and 2. Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue marbles we have by the total marbles in the jar: 18/45. We will keep it in this form because this is the standard way of representing chances in stats classes. Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of drawing a blue marble now become: 17/45. To combine these two chances we now multiply them together: 18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and I\'m sure you could do that part on your own. Solution In probability we have two types of events: Independent: Events who\'s probability is not affected by the previous event(s). Dependent: Events who\'s probability can be affected by preious event(s). What we have in this case is a dependent event because when we don\'t replace the marble, it changes the amount of total marbles inside the jar and thus the chances of drawing another marble changes. Because one event relys on the other we must find their probabilities separately at first. So what we do is figure out the probability of pulling out a blue marble and then the probability of pulling out another after not replacing the first. We will call these event 1 and 2. Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue marbles we have by the total marbles in the jar: 18/45. We will keep it in this form because this is the standard way of representing chances in stats classes. Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of drawing a blue marble now become: 17/45. To combine these two chances we now multiply them together: 18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and I\'m sure you could do that part on your own..

In probability we have two types of eventsIndependent Events who.pdf

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i have written ths code as per your requirements with clear comments as shown below with output package com.sanfoundry.hardgraph; import java.util.Scanner; import java.util.Vector; class Node1 { int name; // node ID, started from 0 to n-1 Vector preds; // predecessors (String) Vector neibs; // neighbors (String) Vector backs; // backward edges -node is end vertex (Integer) Vector fors; // forward edges -node is start vertex (Integer) int pNode; // previous node on the augmenting path int pEdge; // from which edge this node comes on the augmenting // path public Node1(int id) { name = id; backs = new Vector(); fors = new Vector(); pNode1 = -1; pEdge = -1; } } class Edge { int name; // edge ID, started from 0 to n-1 int start; // start vertex of this edge int end; // end vertex of this edge int direct; // forwards (+1) or backwards (-1) on augmenting path // if 0 then not part of augmenting path int capacity; // capacity int flow; // current flow public Edge(int id) { name = id; start = -1; end = -1; direct = 0; // default is neither capacity = 0; flow = 0; } public String toString() { return name + \": s=\" + start + \" e=\" + end + \" d=\" + direct; } } public class LongestPathinDAG { int n; // number of nodes int target; // destination node int minLength; // the minimal length of each path Node1[] v; // used to store Nodes Edge[] e; // used to store Edges int[] path; // used to store temporary path int length = 0; // length of the path int distance = 0; // distance of the path int[] bestPath; // used to store temporary path int bestLength = 0; // length of the longest path int bestDistance = -1000000; // distance of the longest path int[] visited; // used to mark a node as visited if set as // 1 public LongestPathinDAG() { Scanner sc = new Scanner(System.in); System.out.println(\"Enter the number of vertices: \"); n = sc.nextInt(); System.out.println(\"Enter the number of edges: \"); int m = sc.nextInt(); v = new Node1[n]; e = new Edge[m]; System.out.println(n + \" nodes and \" + m + \" edges.\"); for (int i = 0; i < n; i++) v[i] = new Node1(i); int i = 0; while (i < e.length) { Edge edge = new Edge(i); int sVal = sc.nextInt(); edge.start = sVal;// sc.nextInt(); int eVal = sc.nextInt(); edge.end = eVal;// sc.nextInt(); edge.capacity = sc.nextInt(); System.out.println(\" edge: \" + edge.start + \" - \" + edge.end + \" : \" + edge.capacity); edge.flow = 0; e[i] = edge; v[sVal].fors.add(i); v[eVal].backs.add(i); i++; if (i == m) break; } visited = new int[v.length]; path = new int[v.length]; bestPath = new int[v.length]; sc.close(); } /* * this function looks for a longest path starting from being to end, * using the backtrack depth-first search. */ public boolean findLongestPath(int begin, int end, int minLen) { /* * compute a longest path from begin to end */ target = end; bestDistance = -100000000; minLength = minLen; dfsLongestPath(begin); if (bestDistance == -100000000) return false; else return true; } private void dfsLongestPath(int curr.

i have written ths code as per your requirements with clear comments.pdf

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Hi, Please fidn the Answer. //////////////////Sorting,h is Header file which contians all Function decalartion #include #include // std::cout #include // std::shuffle #include // std::array #include // std::default_random_engine #include using namespace std; class Sorting { private: int num; public: Sorting::Sorting() { }; ///Buble Sort Input array and n Number of Elemnets void BubbleSorting(int array[], int n); void SelectionSorting(int array[], int n); void ShuffleArray(int array[], int n); void DisplayArray(int array[], int n); void displayMenu(); int choice; }; //////////////////////////Sorting.cpp is class fuction Which Implemnets the Function #include \"Sorting.h\" #include using namespace std; ///Sorting the Array Using Bubble Sort... void Sorting::BubbleSorting(int Array[],int n) { int temp = 0; cout << \"\ ------------ BUBBLE SORT ------------ \ \ \"; for (int i = 1; iArray[j + 1]) { temp = Array[j]; Array[j] = Array[j + 1]; Array[j + 1] = temp; } } cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \"; DisplayArray(Array, n); } //////SelectionSorting input Array and n is number of Elemnets////////////////////// void Sorting::SelectionSorting(int Array[], int n) { int i, n, p, k, min, loc, temp; for (p = 1; p <= n - 1; p++) // Loop for Pass { min = Array[p]; // Element Selection loc = p; for (k = p + 1; k <= n; k++) // Finding Min Value { if (min > Array[k]) { min = Array[k]; loc = k; } } temp = Array[p]; // Swap Selected Element and Min Value Array[p] = Array[loc]; Array[loc] = temp; } cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \"; DisplayArray(Array, n); } //////SelectionSorting input Array and n is number of Elemnets////////////////////// ///exchange each element with a randomly chosen element. It\'s possible that an element will be exchanged with itself, but there is no problem with that void Sorting::ShuffleArray(int Array[], int n) { cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \"; for (int i = 0; i #include \"Sorting.h\" using namespace std; int main() { bool SelectionOn = true; Sorting Sorting; int choice; int n = 10; int *Arry; while (SelectionOn != false) { cout << \"*******************************\ \"; cout << \" 1 - Enter User Input Array of Elements.\ \"; cout << \" 2 - Display User Input.\ \"; cout << \" 3 - Shuffle User Input the Elements i.\ \"; cout << \" 4 - Bubble Sorting.\ \"; cout << \" 5 - Selection Sorting.\ \"; cout << \" 6 - Exit.\ \"; cout << \" 7 - Display Menu Again. \ \"; cout << \" Enter your choice and press return: \"; cin >> choice; switch (choice) { case 1: cout << \"Enter User Input Array of Elements\ \"; cout << \"Enter No. of Elements=\"; cin >> n; Arry = new int[n]; cout << \"\ Enter Elements=\ \"; for (int i = 1; i <= n; i++) { cin >> Arry[i]; } break; case 2: cout << \"DISPLAY OF ARRAY\ \"; if (Arry != NULL) { Sorting.DisplayArray(Arry, n); } else { cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \"; } break; case 3: cout << \"SHUFFLE OF A.

Hi,Please fidn the Answer.Sorting,h is Header .pdf

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Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and central North America. It is found in mesophytichardwood forests from Nova Scotia west to Minnesota, south to northern Florida, and southwest to eastern Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana Solution Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and central North America. It is found in mesophytichardwood forests from Nova Scotia west to Minnesota, south to northern Florida, and southwest to eastern Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana.

Fraxinus americana (white ash or American ash) is a species of ash t.pdf

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D ) Case IVSolutionD ) Case IV.pdf

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AnswerD. they cannot reproduce in the hostSubunits of vaccine co.pdf

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Ans.1 There are four different position values: •static •relative •fixed •absolute (a) Static: This is default position. And they are not affected by any tag properties like top,left,bottom and right. Ex: Hello this is Static Explanation: The property left:50px will not do any affect to this p tag, as said it just displays as normal flow. (b)Relative An element with position: relative; is positioned relative to its normal position. That is if theres any other element above this element, then it wont effect or cover that above element. Ex: Please dont come here! Hello this is relative Explanation: Give any top,left,right or bottom value to the p tag it wont affect the h1 tag above it, instead it will arrange itself at a position relative to that h1 tag. (c)Fixed This is somewhat powerful than static and relative. An element with position: fixed; is positioned relative to the viewport. Ex: Hey you will hide me for sure. Im fixed element! Explanation: The line in p tag is positioned to top:0px, so the element above p tag which is h1 tag will be affected by this p tag. Inorder to put at right place you need to position it carefully. (d)Absolute This is somewhat similar to fixed and relative. Here the element is positioned(or fixed) according to the last element.And can\'t go top of its top element. Ex: Hi I won\'t be coveredHello I will be covered Hi Im absolute element. Explanation: The p tag which is \'absolute\' positioned to top= 0px will affect its tag above it, that is h2 tag. But it wont affect h1 tag, so you need to position accordingly. Ans.2 In the HTML5 standard, the <\"article\"> element defines a complete, self-contained block of related elements. The <\"section\"> element is defined as a block of related elements. The article element got headers. Ex: Ans.3 HTML5 semantic elements: tag Description Defines an article Defines content aside from the page content Defines additional details that the user can view or hide Defines a caption for a element Specifies self-contained content, like illustrations, diagrams, photos, code listings, etc. Defines a footer for a document or section Specifies a header for a document or section Specifies the main content of a document Defines marked/highlighted text Defines navigation links Defines a section in a document Defines a visible heading for a element Defines a date/time Ans.4 a. #myId, .myClass p, #myidTwo p:first-child a { color;blue; } b. #myId { color:blue; } .myClass p { color:blue; } #myidTwo p:first-child a { color:blue; } Explanation: Actually both (a) and (b) do same job that is putting blue color property, but they are implemented in different ways: In (a) it includes style property defined on set of classes and ids. So any property defined inside it will be applied to all classes/ids(#myId, .myClass p, #myidTwo p:first-child a).This is useful when you are going to give same colour to some set of elements . In (b) it defines css style properties for each class/id separately. It is use.

Ans.1 There are four different position values •static •relat.pdf

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Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid spores by meiosis, gametophyte produces haploid gametes by mitosis. Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows travelling to different areas Solution Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid spores by meiosis, gametophyte produces haploid gametes by mitosis. Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows travelling to different areas.

Ancestral features - multicellular , photosynthetic, cell contains .pdf

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all frangmets...dna ligase joins...cells containing...white co.pdf

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Access: It is a security control who are taking permission to view or use the information that is there inside. Asset: It is a data or device or component which secures the data related activities. Attack: It is an attempt to altering or taking the data without any permission. Control safeguard or countermeasure: It is a process of minimizing the attacks by using some techniques is called counter measure. exploit: is an attack on a computer system, especially one that takes advantage of a particular vulnerability that the system offers to intruders Exposure: It is defined as exposing the information without the knowledge of the lead of theat information. Loss: It says when we loose any data this keyword works. protection profile or security posture: it is typically created by a user or user community and provides an implementation independent specification of information assurance security requirements. Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a vulnerability. sujects and objects: A subject is usually a human user or process running in memory. An object is any resource that exists anywhere a subject can access it (in memory, on disk, across a communications channel, in \"the cloud\", etc.). Threat – Anything that can exploit a vulnerability, intentionally or accidentally, and obtain, damage, or destroy an asset. Threat agent: Method used in breaching the security of a facility, operation, or system by exploiting a vulnerability. Vulnerability – Weaknesses or gaps in a security program that can be exploited by threats to gain unauthorized access to an asset. Solution Access: It is a security control who are taking permission to view or use the information that is there inside. Asset: It is a data or device or component which secures the data related activities. Attack: It is an attempt to altering or taking the data without any permission. Control safeguard or countermeasure: It is a process of minimizing the attacks by using some techniques is called counter measure. exploit: is an attack on a computer system, especially one that takes advantage of a particular vulnerability that the system offers to intruders Exposure: It is defined as exposing the information without the knowledge of the lead of theat information. Loss: It says when we loose any data this keyword works. protection profile or security posture: it is typically created by a user or user community and provides an implementation independent specification of information assurance security requirements. Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a vulnerability. sujects and objects: A subject is usually a human user or process running in memory. An object is any resource that exists anywhere a subject can access it (in memory, on disk, across a communications channel, in \"the cloud\", etc.). Threat – Anything that can exploit a vulnerability, intentionally or accide.

Access It is a security control who are taking permission to view o.pdf

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9) Answer: import java.util.*; import java.lang.*; import java.io.*; class Test { public static void main(String args[]) { int revenue; int expenses; Scanner in = new Scanner(System.in); System.out.println(\"Enter revenue\"); revenue=in.nextInt(); System.out.println(\"Enter expenses\"); expenses=in.nextInt(); if( revenue > expenses ) { System.out.print(\"in the black\"); }else { System.out.print(\"in the red\"); } } } Solution 9) Answer: import java.util.*; import java.lang.*; import java.io.*; class Test { public static void main(String args[]) { int revenue; int expenses; Scanner in = new Scanner(System.in); System.out.println(\"Enter revenue\"); revenue=in.nextInt(); System.out.println(\"Enter expenses\"); expenses=in.nextInt(); if( revenue > expenses ) { System.out.print(\"in the black\"); }else { System.out.print(\"in the red\"); } } }.

9)Answerimport java.util.; import java.lang.; import java.pdf

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Transport (British English) or Transportation (American English) ransportation has developed along three basic Mode (Media):- 1. Land Transportation (way)– (a) Road Transportation (b) Rail Transportation 2. Water Transportation 3. Air Transportation Tramway Inland water transport Ocean transport These may be classified as under: (a). Liners (b). Tramps Liners Vs Tramps Figure- Layout airport runway design TRAFFIC SIGNS

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The Department of Emergency Medicine at Carolinas Medical Center is passionate about education! Dr. Michael Gibbs is a world-renowned clinician and educator and has helped guide numerous young clinicians on the long path of Mastery of Emergency Medical Care. With his oversight, the EMGuideWire team aim to help augment our understanding of emergent imaging. You can follow along with the EMGuideWire.com team as they post these educational, self-guided radiology slides or you can also use this section to learn more in-depth about specific conditions and diseases. This Radiology Reading Room pertains to Sternal Fractures and Dislocations and is brought to you by Carrie Bissell, MD, Aaron Fox, MD, Kendrick Lim, MD, Stephanie Jensen, MD, and Olivia Rice, MD. It is has special guest editor: Sean Dieffenbaugher, MD and Laurence Kempton, MD

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