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The three major forms of business organizations are: 1. Sole Proprietorship 2. Partnership & 3. Corporation Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and responsibilities are with the sole proprietor of the organisation Partnership: Ownership: In a partnership form of organisation more than one person come together to form a business and all of them bring in capital in a decided proportion and they own such decided portion of the organisation. They share liabilities and responsibilities in the pre determined proportion i.e. profit sharing ratio It is managed by all the partners together or by any working partner as decided by all of the partners. Corporation : Many people contribute to the Capital of the organisation and the management lies with the directors and the management ho are elected by the shareholders and are completely answerable and responsible to all the stakeholders. Solution The three major forms of business organizations are: 1. Sole Proprietorship 2. Partnership & 3. Corporation Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and responsibilities are with the sole proprietor of the organisation Partnership: Ownership: In a partnership form of organisation more than one person come together to form a business and all of them bring in capital in a decided proportion and they own such decided portion of the organisation. They share liabilities and responsibilities in the pre determined proportion i.e. profit sharing ratio It is managed by all the partners together or by any working partner as decided by all of the partners. Corporation : Many people contribute to the Capital of the organisation and the management lies with the directors and the management ho are elected by the shareholders and are completely answerable and responsible to all the stakeholders..

The three major forms of business organizations are1. Sole Propri.pdf

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the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging, implementations like TCP/IP conform to the mode Solution the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging, implementations like TCP/IP conform to the mode.

the OSI model is an idea. it is abstract it has no value without imp.pdf

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The EVA metric effectively measures the amount of shareholder wealth that the firm\'s management has added to the firm\'s value during a time period. If EVA is positive , then management has increased value. While there cannot be a single factor which can be attributed to the managers performing their primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both aim towards creation of value. The only difference is in the calculation and the interpretation as defined by mathematical paradigms. EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This indicates the wealth the company generated over and above the cost of capital. THe MVA on the other hand is simply the difference between the current total market value and the capital contributed by investors (both shareholders and bondholders). Thus as the company performs well on the EVA metric and as it retained earnings rise so will the business. And as the business rises so will the share price and hence the MVA. SO basically it is all interrelated. EBIT = 80M-52M = 28M EVA = 28* (1-0.4) - 115*0.125 = 2.425 M MVA of a company is equal to the net present value of all future EVAs. Solution The EVA metric effectively measures the amount of shareholder wealth that the firm\'s management has added to the firm\'s value during a time period. If EVA is positive , then management has increased value. While there cannot be a single factor which can be attributed to the managers performing their primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both aim towards creation of value. The only difference is in the calculation and the interpretation as defined by mathematical paradigms. EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This indicates the wealth the company generated over and above the cost of capital. THe MVA on the other hand is simply the difference between the current total market value and the capital contributed by investors (both shareholders and bondholders). Thus as the company performs well on the EVA metric and as it retained earnings rise so will the business. And as the business rises so will the share price and hence the MVA. SO basically it is all interrelated. EBIT = 80M-52M = 28M EVA = 28* (1-0.4) - 115*0.125 = 2.425 M MVA of a company is equal to the net present value of all future EVAs..

The EVA metric effectively measures the amount of shareholder wealth.pdf

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Reflection about the centre of the pentagon is not its symmetric and.pdf

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Question not visible. Please state again.SolutionQuestion not .pdf

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Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf

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Part A: Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the disease, only if both his/her parents are carriers. Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\' According to HW equilibrium, 2pq= 1/300 Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele cannot be more than the dominant one. When two carriers cross, 25% chance is there of having the genotype \'tt\'. So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of (1/300 * 1/300) Part B. If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30 If the other parent is normal, then chances of being a carrier is 1/300 When two carriers cross, chances of having homozygous recessive genotype = 25% So, 25% chance of having an affected child is there i.e 25% of (1/30 * 1/300) Part C: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Part D: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Solution Part A: Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the disease, only if both his/her parents are carriers. Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\' According to HW equilibrium, 2pq= 1/300 Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele cannot be more than the dominant one. When two carriers cross, 25% chance is there of having the genotype \'tt\'. So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of (1/300 * 1/300) Part B. If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30 If the other parent is normal, then chances of being a carrier is 1/300 When two carriers cross, chances of having homozygous recessive genotype = 25% So, 25% chance of having an affected child is there i.e 25% of (1/30 * 1/300) Part C: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30) Part D: Again 25% chance of having an affected child is there. You can also write as: 25% of (1/30 * 1/30).

Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf

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null is a subset of every setTrueSolutionnull is a sub.pdf

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LiOH Sol.pdf

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Information is a valuable asset that can make or break your business. When properly managed it allows you to operate with confidence. Information security management gives you the freedom to grow, innovate and broaden your customer-base in the knowledge that all your confidential information will remain that way. In this information era, Security has become more difficult to define and enforce. information security was limited to controlling physical access to oral or written communications. The importance of information security led societies to develop innovative ways of protecting their information. Recent innovations in information technology, like the Internet, have made it possible to send vast quantities of data across the globe with ease. However, the challenge of controlling and protecting that information has grown exponentially now that data can be easily transmitted, stored, copied, manipulated, and destroyed. Within a large organization information technology generally refers to laptop and desktop computers, servers, routers, and switches that form a computer network, although information technology also includes fax machines, phone and voice mail systems, cellular phones, and other electronic systems. A growing reliance on computers to work and communicate has made the control of computer networks an important part of information security. Unauthorized access to paper documents or phone conversations is still an information security concern, but the real challenge has become protecting the security of computer networks, especially when they are connected to the Internet. Most large organizations have their own local computer network, or intranet, that links their computers together to share resources and support the communications of employees and others with a legitimate need for access. Almost all of these networks are connected to the Internet and allow employees to go “online.” Solution Information is a valuable asset that can make or break your business. When properly managed it allows you to operate with confidence. Information security management gives you the freedom to grow, innovate and broaden your customer-base in the knowledge that all your confidential information will remain that way. In this information era, Security has become more difficult to define and enforce. information security was limited to controlling physical access to oral or written communications. The importance of information security led societies to develop innovative ways of protecting their information. Recent innovations in information technology, like the Internet, have made it possible to send vast quantities of data across the globe with ease. However, the challenge of controlling and protecting that information has grown exponentially now that data can be easily transmitted, stored, copied, manipulated, and destroyed. Within a large organization information technology generally refers to laptop and desktop computers, servers, routers, and.

Information is a valuable asset that can make or break your business.pdf

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H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf

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First of all we shoud understand what is HIV / AIDS? HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for increase their life span. If you are HIV-positive, good nutrition can have several benefits. It can: Basic principle of nutrition : So overall the HIV infected patient shuld have high nutritious food havig good protein, carbohydrates,vitamins , minerals to get good immune system to defence by the desease. Thanks all the best Solution First of all we shoud understand what is HIV / AIDS? HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for increase their life span. If you are HIV-positive, good nutrition can have several benefits. It can: Basic principle of nutrition : So overall the HIV infected patient shuld have high nutritious food havig good protein, carbohydrates,vitamins , minerals to get good immune system to defence by the desease. Thanks all the best.

First of all we shoud understand what is HIV AIDSHIV AIDS is .pdf

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diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf

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Dear,The answer is .2.SolutionDear,The answer is .2..pdf

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d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf

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Considering the general solution of the Laplace Equation we get Xn(x) = sin(nx) Yn(y) = Ancosh(ny) + Bnsinh(ny) Lets have Yn(0) = 1 and An = 1 for convienience. As we know Yn() = 0 , hence solving for Bn => Bn = -cosh(n)/sinh(n) => Yn(y) = sinh(n - ny)/sinh(n) Since f(x) = sin(4x) => u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) ) Solution Considering the general solution of the Laplace Equation we get Xn(x) = sin(nx) Yn(y) = Ancosh(ny) + Bnsinh(ny) Lets have Yn(0) = 1 and An = 1 for convienience. As we know Yn() = 0 , hence solving for Bn => Bn = -cosh(n)/sinh(n) => Yn(y) = sinh(n - ny)/sinh(n) Since f(x) = sin(4x) => u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) ).

Considering the general solution of the Laplace Equation we getXn(.pdf

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Cranial nerves which do not innervate the structures of the head but.pdf

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Code Include libraries. import javax.swing.JOptionPane;.pdf

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C code: #include int main(void) { FILE *infp, *outfp; int ID,day,hr,i=1; int Temp_ID,Total_hr; int Line_No=0; if ((infp = fopen(\"emp.txt\", \"r\"))==NULL) { printf(\"Input file cannot be opened\ \"); return-1; } if ((outfp = fopen(\"out.txt\", \"w\"))==NULL) { printf(\"Output file cannot be opened\ \"); return-1; } while(!feof(infp)) { Line_No=Line_No+1; fscanf(infp,\"%d %d\",&ID,&hr); if(Line_No==1) { Temp_ID=ID;day=1;Total_hr=hr; } else { if(Temp_ID==ID) { day=day+1; Total_hr=Total_hr+hr; } else { fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr); Temp_ID=ID;day=1;Total_hr=hr; } } } fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr); fclose(infp); fclose(outfp); } emp.txt: 11 2 11 8 11 5 15 10 18 4 18 16 20 23 20 11 out.txt: 11 3 15 15 1 10 18 2 20 20 2 34 Solution C code: #include int main(void) { FILE *infp, *outfp; int ID,day,hr,i=1; int Temp_ID,Total_hr; int Line_No=0; if ((infp = fopen(\"emp.txt\", \"r\"))==NULL) { printf(\"Input file cannot be opened\ \"); return-1; } if ((outfp = fopen(\"out.txt\", \"w\"))==NULL) { printf(\"Output file cannot be opened\ \"); return-1; } while(!feof(infp)) { Line_No=Line_No+1; fscanf(infp,\"%d %d\",&ID,&hr); if(Line_No==1) { Temp_ID=ID;day=1;Total_hr=hr; } else { if(Temp_ID==ID) { day=day+1; Total_hr=Total_hr+hr; } else { fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr); Temp_ID=ID;day=1;Total_hr=hr; } } } fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr); fclose(infp); fclose(outfp); } emp.txt: 11 2 11 8 11 5 15 10 18 4 18 16 20 23 20 11 out.txt: 11 3 15 15 1 10 18 2 20 20 2 34.

C code#include stdio.hint main(void) { FILE infp, o.pdf

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BasicPizza.java public class BasicPizza { //Declaring instance variables String type; String crust; String ingredients; double cost; //Default Constructor public BasicPizza() { super(); } //Parameterized constructor public BasicPizza(String type) { super(); this.type = type; } //Setters and Getters public String getType() { return type; } public void setType(String type) { this.type = type; } public String getCrust() { return crust; } public void setCrust(String crust) { this.crust = crust; } public String getIngredients() { return ingredients; } public void setIngredients(String ingredients) { this.ingredients = ingredients; } public double getCost() { return cost; } public void setCost() { this.cost = 5; } //toString() method is used to display the contents of the Object.inside it. @Override public String toString() { return \"BasicPizza [type=\" + type + \", crust=\" + crust + \", ingredients=\" + ingredients + \", cost=\" + cost + \"]\"; } } ________________________________________________________________ LiFiCheese.java public class LiFiCheese extends BasicPizza{ //Declaring instance variables private String crust; private double cost; private String ingredients; //Default constructor public LiFiCheese() { super(\"Cheese\"); this.cost=5; } //Setters and getters public void setCrust() { this.crust=\"thin\"; } public String getCrust() { return crust; } public double getCost() { return cost; } public void setCost(double cost) { } public String getIngredients() { return ingredients; } public void setIngredients(String ingredients) { this.ingredients = ingredients; } //toString() method is used to display the contents of the Object.inside it. @Override public String toString() { System.out.println(\"You Ordered :\"); System.out.println(getType()); setCrust(); System.out.println(getCrust()); System.out.println(\"Total Cost of :\"+getCost()); return \"\"; } } __________________________________________________________________________ LiFiPizza.java import java.util.Scanner; public class LiFiPizza extends BasicPizza { //Declaring instance variables private String type; private double cost; private String crust; private String ingredients; //Default constructor public LiFiPizza() { super(); this.type = \"Meat\"; this.cost=5; } //Setters and getters public String getType() { return type; } public void setType(String type) { this.type = type; } public double getCost() { return cost; } public void setCost() { this.cost = this.cost+2; } public String getCrust() { return crust; } public void setCrust(String crust) { if(crust.equals(\"Thin\")) { this.crust=\"Thin\"; } else if(crust.equals(\"Thick\")) { this.crust=\"Thick\"; } } public String getIngredients() { return ingredients; } public void setIngredients(String ingredients) { this.ingredients = ingredients; } //toString() method is used to display the contents of the Object.inside it. @Override public String toString() { System.out.println(\"You Ordered :\"); System.out.println(getType()); System.ou.

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