Calculate the implied stock price assuming an EBITDA multiple of 11..pdf

•

0 likes•11 views

Calculate the implied stock price assuming an EBITDA multiple of 11.2x. (All data in millions) Select one: a. $33.42 b. $41.93 c. $24.14 d. $20.58 e. $21.72Short-term debt:$22.5Shares Outstanding:18.330Convertible debt$0.0Diluted shares outstanding:21.501Long-term debt:$253.5Cash & equivalents:$46.5Total liabilities:$676.8Minority interest:$0.0EBITDA:$60.0Net income:$21.2 Solution EBITDA 60 11.2 672 Enterprise Value 672 Less : Short-Term debt -22.5 Long- term debt -253.5 Add: Cash 46.5 Equity Value 442.5 Diluted shares outstanding 21.501 Implied value of stock 20.58.

Education

More from birajdar2

Exercise 7. Show that if C/(0) (the punctured plane) and U C are conformally equivalent, then C/U has no interior, i.e. the complement of U contains no open sets. Solution Let P be the punctured plane and U be conformally equivalent. Then the complements P\' (of P) and U\'( of U) will be topologically equivalent. But P\' ={0} is a singleton...has no interior points..(hence contains no open sets) So the same it true of U\'. Hence the result.

Exercise 7. Show that if C(0) (the punctured plane) and U C are conf.pdf

birajdar2

Consider the following segment table: What are the physical addresses for the following virtual addresses? 4.44 1.22 Briefly compare segmentation to paging. Solution a) Physical address for 4,44: 2200+44 = 2244 Physical address for 1,22: Illegal reference b) Paging and segmentation are different in the following ways: 1) A page is a contiguous range of memory addresses which is mapped to physical memory whereas a segment is an independent address space. 2) A has only one linear address space whereas a segment has many address spaces. 3) A page is a physical unit whereas a segment is a logical unit..

Consider the following segment table What are the physical addresse.pdf

birajdar2

Can you explain the movement of ions and ion channel activity during the resting and action potential? Solution Action potential is the potential due to the which the impulse is passed from one neuron to another. Resting stage - in resting stage the inside has higher concentration of Potassium ions ie (-ve) sign,wheareas the outside it has high concentration of Sodium ions ie (+ve) ions..!! The membrane is slightly permeable to Sodium ions to transport the Sodium outside maintaining the concentration outside and inside.. Action potential - during this this the membrane becomes much more permeable to Sodium ions which allows the huge of Sodium inside the neuron in a short time making the charge opposite direction with comparison to outside.. The Potassium ions moves outside of the neuron thus creating action potential....

Can you explain the movement of ions and ion channel activity during.pdf

birajdar2

Below is the graph of a polynomial function f with real coefficients.pdf

birajdar2

Are higher than average sea surface temperatures associated with a greater than average number of hurricanes each year? In a 2010 study, the average global sea surface temperature anomaly was recorded, along with the number of Atlantic hurricanes above or below the average for that year. The response variable is: Solution Variables of interest in an experiment (those that are measured or observed) are called response or dependent variable The response variable is: average number of hurricanes each year.

Are higher than average sea surface temperatures associated with a g.pdf

birajdar2

A table of values of an increasing function F is shown. Use the table to find to find lower and upper estimates for integral^30 _10 (x) dx. Solution function is increasing x =4 lower estimate =x[f(10)+f(14)+f(18)+f(22)+f(26)] lower estimate =4*[-12-6-2+1+3] lower estimate =-64 ============================ upper estimate =x[f(14)+f(18)+f(22)+f(26)+f(30)] upper estimate =4*[-6-2+1+3+8] upper estimate =16.

A table of values of an increasing function F is shown. Use the table.pdf

birajdar2

A polygenic trait is determined by a single gene with many different.pdf

birajdar2

4. Phil is conducting a seed germination experiment. He places 3 groups of lettuce seeds in a 34º Celsius incubator with adequate moisture. One set of seeds is placed in a dark area with no light source. A second set is placed under artificial light and third set of seeds is placed in direct sunlight. This experiment is intended to test Phil\'s hypothesis that light is necessary for lettuce seed germination. Based on the experimental design, which group was the control? A. Moisture B. Seeds in the dark C. Type of light D. Temperature E. Germination rate Solution In this study seeds in the dark is control group. The control group does not receive any treatment by the researchers in an experiment and this control group is used for the comparison with other test group. In this experiment seeds in the dark is not receiving any treatment as it is placed in a dark area with no light source. Hence seeds in the dark is control group in this experiment..

4. Phil is conducting a seed germination experiment. He places 3 gro.pdf

birajdar2

21. What is the relationship between the maximum size of aggregates and concrete strengths for the same water-cement ratio 22. Certain harmful materias in aggregates, such as such as some shales, will cause popouts. What are the mechanisms of the popouts? 23. What are the minerals that are considered low shrinkage aggregates? 24. List three negative effects of organic impurities in concrete Solution 21) Relationship between maximum size of aggregates and concrete compreesive strength: Compressive strength of concrete depends on the aggreagte strength not necessarily on the aggragate size. However, with an increase in size of the aggreagte to the maximum the consumption of water-cement ratio decreases, thereby the density of concrete increases. With increase in density, the compressive strength increases. The surface texture of aggregate can either be smooth or rough. A smooth surface can improve the workability, yet a rougher surface generates a stronger bond between the paste and the agggregates creating a higher strength. Angular and elongated aggregates have higher surface to volume ratio, better bond characteristics but require more cement paste to produce a workable mixture. So for the same water-cement ratio, smooth aggregates are preferred to produce higher compressive strength. 22) Mechanisms of popouts: Shales have been found to be montmorillonitic. Popouts could be attributed to the pressures generated by the absorption of water by the monmorillonite. The reaction between cement alkalis and the reactive shale particles form an alkali-silica gel at the surface of the reacted particle. The quantity of gel produced from a single reactive particle, and the rate at which it forms, will be the greatest if the particles are exposed to high-alkali concentrations. Once formed, the alkali silica gel absorbs water and start to swell. The swollen gel occupies the available pore space in the concrete creating pressures, which often causes popouts. 23) Minerals that are considered low shrinkage aggregates: Metakalinite: It is a highly reactive material which is formed as a result of roasting of natural kaolinite at temperatures of 700oC to 900oC. Zeolite: The source of zeolite formation is due to settlement of volcanic ash and then reaction with lake salts. Fly ash: It is a by-product obtained during combustion of coal in thermal power plants. Silica fumes: It is a by-product of the production of elemental silicon or alloys containing silicon, also known as condensed silica fume or microsilica. Slag: Blast furnace slag is a by product of extraction of iron ore. Amongst the mineral admixtures, slag possesses the highest specific gravity( around 2.8 to 3.0). Rice hush ash: It is a highly reactive pozzolan obtained by controlled calcination of rice husks. 24) Negative effects of organic impurities: - Organic impurities derived from decaying vegetation delay the setting time and hardening of concrete. - Silt, clay and dust may form a coating on the a.

21. What is the relationship between the maximum size of aggregates a.pdf

birajdar2

Which of the following are organizer molecules in the avian PMZ? is a multiple answer question. pick all that fit. Chordin Cerebrus Frzb Goosecoid Noggin Follistatin Veg T Vg1 Nodal Sonic hedgehog Solution 1) Cordin 2) Goosecoid 3) VegT 4) Vg1 These are the avian organizer molecules. In summary, there is evidence for an active Wnt/Beta catenin pathway within the vegetal region of the chick embryo, which is radially distributed around the epiblast. In contrast, the Tbx6L/cVg1/CMIX system is restricted to the posterior vegetal region of the embryo. The activities of the two pathways overlap spatially in the posterior marginal zone(PMZ), which thus becomes a focal structure of early avian development..

Which of the following are organizer molecules in the avian PMZ is a.pdf

birajdar2

What are the five stages of team development? Describe each stage and how that might appear in a healthcare setting. Solution What are the five stages of team development? Describe each stage and how that might appear in a healthcare setting. Answer : Team Development : Improvement of the efficiency of work team, by developing objective & function amplification and interpersonal processes. Following are the stages of Team development. Step 1: Forming : Team Formation as per the skills Step 2: Storming : Discussions on views , thoughts , ideas of each team member Step 3: Norming : To decide the work & to take a decision Step 4: Performing : Here actual work starts Step 5: Adjourning :This is the end step of a project , where team leader ensure about the completeness of a project. -------------------------------------------------------------------------------------------------------------------- -------------------------------------------------- Application Team Development in Health care sector : 5 to 6 years back the issues of Swine Flue disease in India was very hot.This diseases was arising in rainy seasons , because of the Wet & moisture atmosphere.At that moment there were no vaccinations available. At that moment this diseases was attacking to so many people. For that disease the famous institute in India , Syram Institute Pvt.Ltd (SIPL), this institute had developed the Vaccination on this Swine Flu disease. For preparing this vaccinations they gone through following stages of team development. Step 1: Forming : SIPL found a team from well background medical history .All were professional executive.They were very much aware about the swine Flu diseases. Step 2: Storming : After forming a team , each & every individuals prospectus is being taken into account. Dissolution , Analysis on thoughts , ideas were done for successful execution. Step 3: Norming : In this step this has been decided to produce vaccinations as a medicine for this diseases. Vaccinations is being designed , tested ,achieved results .Finally hand over for mass productions. Step 4: Performing : In this stage millions of vaccinations are produced and made available time to time for patients within time. Step 5: Adjourning : They completed this project successfully & made available this project to market in next year.At the same time they are maintaining the changes in vaccinations time to time..

What are the five stages of team development Describe each stage an.pdf

birajdar2

What kinds of molecules can be used as metabolic fuel to produce ATP.pdf

birajdar2

1. Choose the best answer for each statment/question. a.) Which statement describes the difference between the normal neuron protein PrPC and the pathogenic version PrPSc In a persistent infection, the host cell is continually releasing new viral particles slowly. In a latent infection, there are periods of time where the virus is not replicating and creating new viral particles. b.) One hypothesis that has been proposed to explain the development of viruses suggests that viruses were beneficial because they allowed for rapid gene transfer, especially in prokaryotic cells, and therefore increased genetic diversity. If most bacteriophages were temperate, then this could have increased horizontal gene transfer between prokaryotic cells and potentially could have increased fitness. With respect to this hypothesis, why would it matter whether the bacteriophages were temperate or virulent? Temperate phages don\'t immediately kill their hosts. c.) One hypothesis that has been proposed to explain the development of viruses suggests that viruses were beneficial because they allowed for rapid gene transfer, especially in prokaryotic cells, and therefore increased genetic diversity. Why is this hypothesis stronger with respect to prokaryotic cells than if large, complex eukaryotic organisms had been involved? In eukaryotic organisms with sexual reproduction, great genetic diversity can be accomplished through crossing over and independent assortment in meiosis. d.) Why do some bacteriophages, like X174, using rolling circle replication instead of the bidirectional replication?Persistent infections always lead to cell lysis, while latent infections never lead to cell lysis.In a latent infection, the host cell is continually releasing new viral particles slowly. In a persistent infection, there are periods of time where the virus is not replicating and creating new viral particles.Persistent infections can cause transformation of the host cell, which can lead to the development of cancer. Latent infections do not cause transformation. In a persistent infection, the host cell is continually releasing new viral particles slowly. In a latent infection, there are periods of time where the virus is not replicating and creating new viral particles. b.) One hypothesis that has been proposed to explain the development of viruses suggests that viruses were beneficial because they allowed for rapid gene transfer, especially in prokaryotic cells, and therefore increased genetic diversity. If most bacteriophages were temperate, then this could have increased horizontal gene transfer between prokaryotic cells and potentially could have increased fitness. With respect to this hypothesis, why would it matter whether the bacteriophages were temperate or virulent?Temperate phages are capable of lysogeny and their genetic material can be integrated into the host genome, causing damage rather than increasing fitness.Virulent phages are capable of lysogeny and their genetic.

1. Choose the best answer for each statmentquestion.a.) Which sta.pdf

birajdar2

1) What does the nonbranching proposal say about split-brain patients? Are there two people present in the body? If so, is either identical with the person who occupied the body before the operation? Solution As far as awareness hypothesis there is a contention that says that the patient has two continuous flows. One half of the globe contending with the other with regards to taking care of an object(s). Inside an ordinary subject contending continuous flows is nothing strange, yet the distinction is by all accounts that the SB understanding can\'t synchronize the two contending streams into a bound together state. In the matter of whether this outcomes in a split identity I don\'t generally know. Yes. In right gave patients, the left-mind stays associated with beon and will show the greater part of the subject\'s unique attributes. (Switched in left gave patients.) from the beginning, the correct cerebrum will show no identity since it is recently curious to see what happens, as some time recently..

1) What does the nonbranching proposal say about split-brain patient.pdf

birajdar2

What are the levels of biodiversity. And how are they valuable? Drag the labels onto the concept map to review the ecological value and practical benefits of biodiversity. First fill in the pink targets using the pink labels. Then complete the rest of the concept map using the blue labels. Solution a. Ecosystem diversity b. Species diversity c. Genetic diversity d . Ecosystem services: Ecosystem diversity provides Ecosystem services including water purification. e. Community interactions: these community interactions lead to emergence of keystone species. f. genes with practical uses g. New medicines h.Species survival.

What are the levels of biodiversity. And how are they valuable Drag.pdf

birajdar2

Two students were debating how to classify superscalar operation versus the use of VLIW instructions. The debate was over which of these is an organizational feature and which is an architectural feature. Indicate which point of view you would support and explain why. Solution A superscalar processor is a CPU that implements a form of parallelism calledinstruction-level parallelism within a single processor. It therefore allows for morethroughput (the number of instructions that can be executed in a unit of time) than would otherwise be possible at a givenclock rate. A superscalar processor can execute more than one instruction during a clock cycle by simultaneously dispatching multiple instructions to different execution units on the processor. Each execution unit is not a separate processor (or a core if the processor is a multi- core processor), but an execution resource within a single CPU such as an arithmetic logic unit. In Flynn\'s taxonomy, a single-core superscalar processor is classified as anSISD processor (Single Instruction stream, Single Data stream), though many superscalar processors support short vector operations and so could be classified as SIMD (Single Instruction stream, Multiple Data streams). A multi-core superscalar processor is classified as an MIMD processor (Multiple Instruction streams, Multiple Data streams). While a superscalar CPU is typically alsopipelined, pipelining and superscalar execution are considered different performance enhancement techniques. The former executes multiple instructions in the same execution unit in parallel by dividing the execution unit into different phases, whereas the latter executes multiple instructions in parallel by using multiple execution units. The superscalar technique is traditionally associated with several identifying characteristics (within a given CPU): Instructions are issued from a sequential instruction stream The CPU dynamically checks for data dependencies between instructions at run time (versus software checking at compile time) The CPU can execute multiple instructions per clock cycle.

Two students were debating how to classify superscalar operation ver.pdf

birajdar2

Throughout the semester, we have been working on command line applications. While the command line can be a powerful form of interacting with the computer, most modern applications have a graphical user interface, or GUI. This program will give you some initial experience making GUI applications with Swing. 1. Write an original GUI application in Java using Swing. The program can be a simple game, or an application that demonstrates some of the capabilities of Swing, for example. 2. The application must do something non-trivial that is controlled by GUI components such as buttons, text boxes, drop-down boxes, etc. For example, it might draw different things, or change colors, based on button presses. 3. The GUI elements should be layed out reasonably using layout managers. 4. At least one element in your GUI program besides the title bar must include your name. Suggestions Get the skeleton of your GUI working first. First get the buttons drawing properly, then go on to listen to and respond to events. Initially, hook up the event listeners so that they print something to the console when a button is pressed. That way you can know that your code is getting events. If your program will draw shapes, make a subclass of JPanel called \"DrawPanel\" or something like that to draw the objects and your name. Override the \"paintComponent\" method of this class. Solution import javax.swing.*; import java.awt.*; class gui{ public static void main(String args[]){ //Creating the Frame JFrame frame = new JFrame(\"Chat Frame\"); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setSize(400,400); //Creating the MenuBar and adding components JMenuBar mb = new JMenuBar(); JMenu m1 = new JMenu(\"FILE\"); JMenu m2 = new JMenu(\"Help\"); mb.add(m1); mb.add(m2); JMenuItem m11 = new JMenuItem(\"Open\"); JMenuItem m22 =new JMenuItem(\"Save as\"); m1.add(m11); m1.add(m22); //Creating the panel at bottom and adding components JPanel panel = new JPanel(); // the panel is not visible in output JLabel label = new JLabel(\"Enter Text\"); JTextField tf = new JTextField(10);// accepts upto 10 characters JButton send = new JButton(\"Send\"); JButton reset = new JButton(\"Reset\"); panel.add(label);// Components Added using Flow Layout panel.add(tf); panel.add(send); panel.add(reset); // Text Area at the Center JTextArea ta = new JTextArea(); //Adding Components to the frame. frame.getContentPane().add(BorderLayout.SOUTH,panel); frame.getContentPane().add(BorderLayout.NORTH,mb); frame.getContentPane().add(BorderLayout.CENTER,ta); frame.setVisible(true); } }.

Throughout the semester, we have been working on command line applic.pdf

birajdar2

the formula to compute the spending multiplier is: Solution Spending multiplier= Change in equilibrium real GDP/ initial change in autonomous spending. This multiplier shows how much economy\'s output can rise with an increase in spending. This change in spending could be change in consumption, investment,govt spending and net exports. Formula for spending multiplier= 1/(1-MPC) or 1/MPS; where MPC is maginal propensity to consume..

the formula to compute the spending multiplier isSolutionS.pdf

birajdar2

The figure illustrates an interaction between a(n) CTL and its target cell. antigen presenting cell and a plasma cell. antigen presenting cell and a B lymphocyte. antigen presenting cell and a T lymphocyte. NK cell and its target cell. Solution TCR (T-cell receptor) and MHC|| are complex to present APC( antigen presenting cell) CD4 CD8 are xo stimulatory signals for activation of T-cell So answer is Interaction between antigen presenting cell and T-lymphocytes.

The figure illustrates an interaction between a(n) CTL and its tar.pdf

birajdar2

The implementation for Mergesort given in Section 7.4 takes an array as input and soils that array. At the beginning of Section 7.4 there is a simple pseudocode implementation for sorting a linked list using Mergesort. Implement both a linked list-based version of Mergesort and the array-based version of Mergesort, and compare and analyze their running times. Solution program import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Iterator; public class Merge { public static int[] merge(int [] list) { if (list.length <= 1) { return list; } int[] first = new int[list.length / 2]; int[] second = new int[list.length - first.length]; System.arraycopy(list, 0, first, 0, first.length); System.arraycopy(list, first.length, second, 0, second.length); merge(first); merge(second); mergesort(first, second, list); return list; } private static void mergesort(int[] first, int[] second, int [] result) { int iFirst = 0; int iSecond = 0; int j = 0; while (iFirst < first.length && iSecond < second.length) { if (first[iFirst] < second[iSecond]) { result[j] = first[iFirst]; iFirst++; } else { result[j] = second[iSecond]; iSecond++; } j++; } System.arraycopy(first, iFirst, result, j, first.length - iFirst); System.arraycopy(second, iSecond, result, j, second.length - iSecond); } public static void main(String args[]) throws Exception { String list=\"\"; int i=0,n=0; Merge s= new Merge(); ArrayList arrlist=new ArrayList(); System.out.println(\" \"); System.out.println(\" \"); System.out.println(\"Please enter the list of elements,one element per line\"); System.out.println(\" write \'STOP\' when list is completed \"); BufferedReader bf=new BufferedReader(new InputStreamReader(System.in)); while(!(list=bf.readLine()).equalsIgnoreCase(\"stop\")) { int intelement=Integer.parseInt(list); arrlist.add(intelement); } int elementlist[] = new int[arrlist.size()]; Iterator iter = arrlist.iterator(); for (int j=0;iter.hasNext();j++) { elementlist[j] = iter.next(); } elementlist=merge(elementlist); System.out.println(\" \"); System.out.println(\" \"); System.out.println(\" \"); System.out.println(\"Values after Merge Sort : \"); for (int j=0;j.

The implementation for Mergesort given in Section 7.4 takes an array .pdf

birajdar2

More from birajdar2 (20)