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- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return you get ion-dipole forces of the Cs(+) and I(-) in the solution - Dispersion Forces and Metallic bonds Solution - You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return you get ion-dipole forces of the Cs(+) and I(-) in the solution - Dispersion Forces and Metallic bonds.
- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
ankitmobileshop235
#include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2); p1->insertFront(\'a\'); p1->print(); system(\"pause\"); return 0; } Solution #include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2.
#includeiostream struct node { char value; struct no.pdf
#includeiostream struct node { char value; struct no.pdf
ankitmobileshop235
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another atom. Solution Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another atom..
Two sp3 orbitals are filled by lone electron pair.pdf
Two sp3 orbitals are filled by lone electron pair.pdf
ankitmobileshop235
X-Intercept is the value of x where cross x axis. Another name is the zeros of the function, when y is zero, the graph cross the x-axis The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to it. Solution X-Intercept is the value of x where cross x axis. Another name is the zeros of the function, when y is zero, the graph cross the x-axis The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to it..
X-Intercept is the value of x where cross x axis. Another name is .pdf
X-Intercept is the value of x where cross x axis. Another name is .pdf
ankitmobileshop235
\"whether or not the viruses\" is not used to classify viruses. Viruses are classified based on the type of genetic material. They have either DNA or RNA but not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA may also be single stranded or double stranded. Based on the presence or absence of envelope they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non enveloped. Solution \"whether or not the viruses\" is not used to classify viruses. Viruses are classified based on the type of genetic material. They have either DNA or RNA but not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA may also be single stranded or double stranded. Based on the presence or absence of envelope they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non enveloped..
whether or not the viruses is not used to classify viruses.Vir.pdf
whether or not the viruses is not used to classify viruses.Vir.pdf
ankitmobileshop235
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones. generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text editor, therefore knowing Vi is important. Unlike Nano, Associate in Nursing easy-to-use terminal text editor, Vi doesn’t hold your hand and supplya listingof keyboard shortcuts on the screen. It’s a modal text editor, Associate in Nursingd it\'seach an insert and command mode. Vi may be a terminal application, therefore you’ll ought tobegin it from a terminal window. Use the vi /path/to/file command to open Associate in Nursing existing file with Vi. The vi /path/to/file command additionally works if the file doesn’t exist yet; Vi canproducea brand new file and write it to the desired location once you save. Command Mode This is what you’ll see once you open a get in vi. it\'ssuch as youwillsimplybeginwriting, however you can’t. Vimay be a modal text editor, and it opens in command mode. making an attempt to kind at this screen canlead tosudden behavior. While in command mode, you\'ll move the pointer around with the arrow keys. Press the x key to delete the characterbeneath the pointer. There area unita spread of alternative delete commands — as an example, writingDD(press the d key twice) deletes a whole line of text. You can choose, copy, cut and paste text in command mode. Position the pointer at the left or right aspect of the text you would liketo repeat and press the v key. Move your pointerto pick out text, so press y to repeatthe chosen text or x to chop it. Position your pointer at the required location and press the p key to stick the text youtraced or cut. Insert Mode Aside from command mode, the opposite mode you wishto understandconcerning is insert mode, thatpermitsyou to insert text in Vi. getting into insert mode is simple once you recognize it exists — simply press the i key once when you’ve positioned the pointer in command mode. beginwriting and Vi can insert the characters youkind into the file instead ofmaking an attempt to interpret them as commands. Once you’re worn out insert mode, press the escape key to come to command mode. Saving and Quitting You can save and quit vi from command mode. First, guarantee you’re in command mode by pressing the escape key (pressing the escape key once morewill nothing if you’re already in command mode.) Type :wq and press enter to put in writing the file to disk and quit vi. you\'lladditionally split this command up — as an example, kind :w and press enter to put in writing the file to disk while not quitting or kind :q to quit vi while not saving the file. Vi won’t allow you to quit if you’ve changed the file since you last saved, howeveryou\'llkind :q! and press enter to ignore this warning. Solution Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones. generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text editor, therefore knowing Vi is .
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
ankitmobileshop235
U.S management is trained to provide the executives their own career path; it provides flexibility to the top executive over the journey. The top executive is set forward to accommodate their own journey with their own set of interest. The management is expecting that the top executives will be of good track record and will somehow contribute to the growth of the company and will contribute solidly to the team and work as a team. The top executive should deliver best quality efforts to the other team members. Management is trying to provide a corporate culture or training to the team members and ill deliver appropriate efforts to the company. The management is also trained towards the culture of the company and is dedicated to provide cultural formal training. The U.S. management is focused on dividing the executive on the basis of common language globally and domestically. To ensure the same the management is trained for providing language courses to the executives for better understanding and training to the organization. Solution U.S management is trained to provide the executives their own career path; it provides flexibility to the top executive over the journey. The top executive is set forward to accommodate their own journey with their own set of interest. The management is expecting that the top executives will be of good track record and will somehow contribute to the growth of the company and will contribute solidly to the team and work as a team. The top executive should deliver best quality efforts to the other team members. Management is trying to provide a corporate culture or training to the team members and ill deliver appropriate efforts to the company. The management is also trained towards the culture of the company and is dedicated to provide cultural formal training. The U.S. management is focused on dividing the executive on the basis of common language globally and domestically. To ensure the same the management is trained for providing language courses to the executives for better understanding and training to the organization..
U.S management is trained to provide the executives their own career.pdf
U.S management is trained to provide the executives their own career.pdf
ankitmobileshop235
There are many operating systems Real-Time Operating System Real-time applications usually are executed on top of a Real-time Operating System (RTOS). Specific scheduling algorithms can be designed. When possible, static cyclic schedules are calculated off-line. Real-time systems are those systems in which the correctness of the system depends not only on the logical result of computation, but also on the time at which the results are produced. RTOS is therefore an operating system that supports real-time applications by providing logically correct result within the deadline required. Basic Structure is similar to regular OS but, in addition, it provides mechanisms to allow real time scheduling of tasks. Though real-time operating systems may or may not increase the speed of execution, they can provide much more precise and predictable timing characteristics than general-purpose OS. A real-time system is defined as a data processing system in which the time interval required to process and respond to inputs is so small that it controls the environment. The time taken by the system to respond to an input and display of required updated information is termed as the response time. So in this method, the response time is very less as compared to online processing. Real-time systems are used when there are rigid time requirements on the operation of a processor or the flow of data and real-time systems can be used as a control device in a dedicated application. A real-time operating system must have well-defined, fixed time constraints, otherwise the system will fail. For example, Scientific experiments, medical imaging systems, industrial control systems, weapon systems, robots, air traffic control systems, etc. Design considerations Designing a proper RTOS architecture needs some delicate decisions. The basic services like process management, inter-process communication, interrupt handling, or process synchronization have to be provided in an efficient manner making use of a very restricted resource budget. Multi-core architectures need special techniques for process management, memory management, and synchronization. The upcoming Wireless Sensor Networks (WSN) generate special demands for RTOS support leading to dedicated solutions. Another special area is given by multimedia applications. Very high data rates have to be supported under (soft) RT constraints. The key difference between general-computing operating systems and real-time operating systems is the need for \" deterministic \" timing behavior in the real-time operating systems. Formally, \"deterministic\" timing means that operating system services consume only known and expected amounts of time. In theory, these service times could be expressed as mathematical formulas. These formulas must be strictly algebraic and not include any random timing components. Random elements in service times could cause random delays in application software and could then make the application randomly .
There are many operating systemsReal-Time Operating SystemReal-t.pdf
There are many operating systemsReal-Time Operating SystemReal-t.pdf
ankitmobileshop235
Recommended
- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return you get ion-dipole forces of the Cs(+) and I(-) in the solution - Dispersion Forces and Metallic bonds Solution - You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return you get ion-dipole forces of the Cs(+) and I(-) in the solution - Dispersion Forces and Metallic bonds.
- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
ankitmobileshop235
#include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2); p1->insertFront(\'a\'); p1->print(); system(\"pause\"); return 0; } Solution #include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2.
#includeiostream struct node { char value; struct no.pdf
#includeiostream struct node { char value; struct no.pdf
ankitmobileshop235
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another atom. Solution Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another atom..
Two sp3 orbitals are filled by lone electron pair.pdf
Two sp3 orbitals are filled by lone electron pair.pdf
ankitmobileshop235
X-Intercept is the value of x where cross x axis. Another name is the zeros of the function, when y is zero, the graph cross the x-axis The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to it. Solution X-Intercept is the value of x where cross x axis. Another name is the zeros of the function, when y is zero, the graph cross the x-axis The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to it..
X-Intercept is the value of x where cross x axis. Another name is .pdf
X-Intercept is the value of x where cross x axis. Another name is .pdf
ankitmobileshop235
\"whether or not the viruses\" is not used to classify viruses. Viruses are classified based on the type of genetic material. They have either DNA or RNA but not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA may also be single stranded or double stranded. Based on the presence or absence of envelope they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non enveloped. Solution \"whether or not the viruses\" is not used to classify viruses. Viruses are classified based on the type of genetic material. They have either DNA or RNA but not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA may also be single stranded or double stranded. Based on the presence or absence of envelope they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non enveloped..
whether or not the viruses is not used to classify viruses.Vir.pdf
whether or not the viruses is not used to classify viruses.Vir.pdf
ankitmobileshop235
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones. generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text editor, therefore knowing Vi is important. Unlike Nano, Associate in Nursing easy-to-use terminal text editor, Vi doesn’t hold your hand and supplya listingof keyboard shortcuts on the screen. It’s a modal text editor, Associate in Nursingd it\'seach an insert and command mode. Vi may be a terminal application, therefore you’ll ought tobegin it from a terminal window. Use the vi /path/to/file command to open Associate in Nursing existing file with Vi. The vi /path/to/file command additionally works if the file doesn’t exist yet; Vi canproducea brand new file and write it to the desired location once you save. Command Mode This is what you’ll see once you open a get in vi. it\'ssuch as youwillsimplybeginwriting, however you can’t. Vimay be a modal text editor, and it opens in command mode. making an attempt to kind at this screen canlead tosudden behavior. While in command mode, you\'ll move the pointer around with the arrow keys. Press the x key to delete the characterbeneath the pointer. There area unita spread of alternative delete commands — as an example, writingDD(press the d key twice) deletes a whole line of text. You can choose, copy, cut and paste text in command mode. Position the pointer at the left or right aspect of the text you would liketo repeat and press the v key. Move your pointerto pick out text, so press y to repeatthe chosen text or x to chop it. Position your pointer at the required location and press the p key to stick the text youtraced or cut. Insert Mode Aside from command mode, the opposite mode you wishto understandconcerning is insert mode, thatpermitsyou to insert text in Vi. getting into insert mode is simple once you recognize it exists — simply press the i key once when you’ve positioned the pointer in command mode. beginwriting and Vi can insert the characters youkind into the file instead ofmaking an attempt to interpret them as commands. Once you’re worn out insert mode, press the escape key to come to command mode. Saving and Quitting You can save and quit vi from command mode. First, guarantee you’re in command mode by pressing the escape key (pressing the escape key once morewill nothing if you’re already in command mode.) Type :wq and press enter to put in writing the file to disk and quit vi. you\'lladditionally split this command up — as an example, kind :w and press enter to put in writing the file to disk while not quitting or kind :q to quit vi while not saving the file. Vi won’t allow you to quit if you’ve changed the file since you last saved, howeveryou\'llkind :q! and press enter to ignore this warning. Solution Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones. generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text editor, therefore knowing Vi is .
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
ankitmobileshop235
U.S management is trained to provide the executives their own career path; it provides flexibility to the top executive over the journey. The top executive is set forward to accommodate their own journey with their own set of interest. The management is expecting that the top executives will be of good track record and will somehow contribute to the growth of the company and will contribute solidly to the team and work as a team. The top executive should deliver best quality efforts to the other team members. Management is trying to provide a corporate culture or training to the team members and ill deliver appropriate efforts to the company. The management is also trained towards the culture of the company and is dedicated to provide cultural formal training. The U.S. management is focused on dividing the executive on the basis of common language globally and domestically. To ensure the same the management is trained for providing language courses to the executives for better understanding and training to the organization. Solution U.S management is trained to provide the executives their own career path; it provides flexibility to the top executive over the journey. The top executive is set forward to accommodate their own journey with their own set of interest. The management is expecting that the top executives will be of good track record and will somehow contribute to the growth of the company and will contribute solidly to the team and work as a team. The top executive should deliver best quality efforts to the other team members. Management is trying to provide a corporate culture or training to the team members and ill deliver appropriate efforts to the company. The management is also trained towards the culture of the company and is dedicated to provide cultural formal training. The U.S. management is focused on dividing the executive on the basis of common language globally and domestically. To ensure the same the management is trained for providing language courses to the executives for better understanding and training to the organization..
U.S management is trained to provide the executives their own career.pdf
U.S management is trained to provide the executives their own career.pdf
ankitmobileshop235
There are many operating systems Real-Time Operating System Real-time applications usually are executed on top of a Real-time Operating System (RTOS). Specific scheduling algorithms can be designed. When possible, static cyclic schedules are calculated off-line. Real-time systems are those systems in which the correctness of the system depends not only on the logical result of computation, but also on the time at which the results are produced. RTOS is therefore an operating system that supports real-time applications by providing logically correct result within the deadline required. Basic Structure is similar to regular OS but, in addition, it provides mechanisms to allow real time scheduling of tasks. Though real-time operating systems may or may not increase the speed of execution, they can provide much more precise and predictable timing characteristics than general-purpose OS. A real-time system is defined as a data processing system in which the time interval required to process and respond to inputs is so small that it controls the environment. The time taken by the system to respond to an input and display of required updated information is termed as the response time. So in this method, the response time is very less as compared to online processing. Real-time systems are used when there are rigid time requirements on the operation of a processor or the flow of data and real-time systems can be used as a control device in a dedicated application. A real-time operating system must have well-defined, fixed time constraints, otherwise the system will fail. For example, Scientific experiments, medical imaging systems, industrial control systems, weapon systems, robots, air traffic control systems, etc. Design considerations Designing a proper RTOS architecture needs some delicate decisions. The basic services like process management, inter-process communication, interrupt handling, or process synchronization have to be provided in an efficient manner making use of a very restricted resource budget. Multi-core architectures need special techniques for process management, memory management, and synchronization. The upcoming Wireless Sensor Networks (WSN) generate special demands for RTOS support leading to dedicated solutions. Another special area is given by multimedia applications. Very high data rates have to be supported under (soft) RT constraints. The key difference between general-computing operating systems and real-time operating systems is the need for \" deterministic \" timing behavior in the real-time operating systems. Formally, \"deterministic\" timing means that operating system services consume only known and expected amounts of time. In theory, these service times could be expressed as mathematical formulas. These formulas must be strictly algebraic and not include any random timing components. Random elements in service times could cause random delays in application software and could then make the application randomly .
There are many operating systemsReal-Time Operating SystemReal-t.pdf
There are many operating systemsReal-Time Operating SystemReal-t.pdf
ankitmobileshop235
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
The three major forms of business organizations are1. Sole Propri.pdf
ankitmobileshop235
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
the OSI model is an idea. it is abstract it has no value without imp.pdf
ankitmobileshop235
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
The EVA metric effectively measures the amount of shareholder wealth.pdf
ankitmobileshop235
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has trivial centre Solution Reflection about the centre of the pentagon is not its symmetric and the symmetry group has trivial centre.
Reflection about the centre of the pentagon is not its symmetric and.pdf
Reflection about the centre of the pentagon is not its symmetric and.pdf
ankitmobileshop235
Question not visible. Please state again. Solution Question not visible. Please state again..
Question not visible. Please state again.SolutionQuestion not .pdf
Question not visible. Please state again.SolutionQuestion not .pdf
ankitmobileshop235
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the halide anion Solution Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the halide anion.
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
ankitmobileshop235
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
Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf
ankitmobileshop235
null is a subset of every set True Solution null is a subset of every set True.
null is a subset of every setTrueSolutionnull is a sub.pdf
null is a subset of every setTrueSolutionnull is a sub.pdf
ankitmobileshop235
LiOH Solution LiOH.
LiOH Sol.pdf
LiOH Sol.pdf
ankitmobileshop235
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
Information is a valuable asset that can make or break your business.pdf
ankitmobileshop235
Ho: there is no relationship between the age of the individual and the individual Solution Ho: there is no relationship between the age of the individual and the individual.
Ho there is no relationship between the age of the individual and t.pdf
Ho there is no relationship between the age of the individual and t.pdf
ankitmobileshop235
H2SO3 - H2O => SO2 The oxide is sulfur dioxide: SO2 Solution H2SO3 - H2O => SO2 The oxide is sulfur dioxide: SO2.
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
ankitmobileshop235
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
First of all we shoud understand what is HIV AIDSHIV AIDS is .pdf
ankitmobileshop235
diffusion rate = 1/M r1/r2 = m2/m1 = t2/t1 10.1/2.16 = m2 /2 molar mass of the unknown gas , m2 = 43.728 g Solution diffusion rate = 1/M r1/r2 = m2/m1 = t2/t1 10.1/2.16 = m2 /2 molar mass of the unknown gas , m2 = 43.728 g.
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
ankitmobileshop235
Dear, The answer is .2. Solution Dear, The answer is .2..
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
ankitmobileshop235
d. Increases venous PO2. systemic venous blood is likely to be 25% saturated with oxygen. PO2 Partial presure of oxygen. Solution d. Increases venous PO2. systemic venous blood is likely to be 25% saturated with oxygen. PO2 Partial presure of oxygen..
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
ankitmobileshop235
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
Considering the general solution of the Laplace Equation we getXn(.pdf
ankitmobileshop235
Cranial nerves which do not innervate the structures of the head but innervate the structures originating from branchial arches are the following Solution Cranial nerves which do not innervate the structures of the head but innervate the structures originating from branchial arches are the following.
Cranial nerves which do not innervate the structures of the head but.pdf
Cranial nerves which do not innervate the structures of the head but.pdf
ankitmobileshop235
Code: //Include libraries. import javax.swing.JOptionPane; //Define a class lPayroll public class lPayroll { //Main function public static void main( String args[] ) { //Initialize variables String lfirst = \"\"; String lLast = \"\"; String lssn = \"\"; String lsalaryString; double lsalary = 0; String lwageString; double lwage = 0; String lhoursString; int lhours = 0; String lsalesString; double lsales = 0; String lrateString; double lrate = 0; String lempTypeString = \"\"; int lempType = 0; int lcount; for(lcount = 0; lcount < 5; lcount++) { //Receive user input lempTypeString = JOptionPane.showInputDialog(null, \"Enter employee type \ \ Enter your choice.\ 1. Salaried\ 2. Hourly\ 3. Commissioned\ 4. Commission plus base\", \"Employee Type\", JOptionPane.QUESTION_MESSAGE); lempType = Integer.parseInt(lempTypeString); switch(lempType) { case 1: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name \", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lsalaryString = JOptionPane.showInputDialog(null, \"Enter weekly salary\", \"lsalary\", JOptionPane.QUESTION_MESSAGE); lsalary = Double.parseDouble(lsalaryString); break; case 2: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lwageString = JOptionPane.showInputDialog(null, \"Enter hourly wage\", \"Hourly lwage\", JOptionPane.QUESTION_MESSAGE); lwage = Double.parseDouble(lwageString); lhoursString = JOptionPane.showInputDialog(null, \"Enter count of hours per week of work\", \"lhours Worked\", JOptionPane.QUESTION_MESSAGE); lhours = Integer.parseInt(lhoursString); break; case 3: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lsalesString = JOptionPane.showInputDialog(null, \"Enter gross sales on weekly basis\", \"Gross lsales\", JOptionPane.QUESTION_MESSAGE); lsales = Double.parseDouble(lsalesString); lrateString = JOptionPane.showInputDialog(null, \"Enter commission percent\", \"Commsission lrate\", JOptionPane.QUESTION_MESSAGE); lrate = Double.parseDouble(lrateString); break; case 4: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_ME.
Code Include libraries. import javax.swing.JOptionPane;.pdf
Code Include libraries. import javax.swing.JOptionPane;.pdf
ankitmobileshop235
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
C code#include stdio.hint main(void) { FILE infp, o.pdf
ankitmobileshop235
https://app.box.com/s/h5mhqoyabotgw05s0df0ltw3e39pgnmy
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
Nguyen Thanh Tu Collection
Presentation on Hindu texts
An overview of the various scriptures in Hinduism
An overview of the various scriptures in Hinduism
Dabee Kamal
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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
The three major forms of business organizations are1. Sole Propri.pdf
ankitmobileshop235
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
the OSI model is an idea. it is abstract it has no value without imp.pdf
ankitmobileshop235
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
The EVA metric effectively measures the amount of shareholder wealth.pdf
ankitmobileshop235
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has trivial centre Solution Reflection about the centre of the pentagon is not its symmetric and the symmetry group has trivial centre.
Reflection about the centre of the pentagon is not its symmetric and.pdf
Reflection about the centre of the pentagon is not its symmetric and.pdf
ankitmobileshop235
Question not visible. Please state again. Solution Question not visible. Please state again..
Question not visible. Please state again.SolutionQuestion not .pdf
Question not visible. Please state again.SolutionQuestion not .pdf
ankitmobileshop235
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the halide anion Solution Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the halide anion.
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
ankitmobileshop235
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
Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf
ankitmobileshop235
null is a subset of every set True Solution null is a subset of every set True.
null is a subset of every setTrueSolutionnull is a sub.pdf
null is a subset of every setTrueSolutionnull is a sub.pdf
ankitmobileshop235
LiOH Solution LiOH.
LiOH Sol.pdf
LiOH Sol.pdf
ankitmobileshop235
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
Information is a valuable asset that can make or break your business.pdf
ankitmobileshop235
Ho: there is no relationship between the age of the individual and the individual Solution Ho: there is no relationship between the age of the individual and the individual.
Ho there is no relationship between the age of the individual and t.pdf
Ho there is no relationship between the age of the individual and t.pdf
ankitmobileshop235
H2SO3 - H2O => SO2 The oxide is sulfur dioxide: SO2 Solution H2SO3 - H2O => SO2 The oxide is sulfur dioxide: SO2.
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
ankitmobileshop235
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
First of all we shoud understand what is HIV AIDSHIV AIDS is .pdf
ankitmobileshop235
diffusion rate = 1/M r1/r2 = m2/m1 = t2/t1 10.1/2.16 = m2 /2 molar mass of the unknown gas , m2 = 43.728 g Solution diffusion rate = 1/M r1/r2 = m2/m1 = t2/t1 10.1/2.16 = m2 /2 molar mass of the unknown gas , m2 = 43.728 g.
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
ankitmobileshop235
Dear, The answer is .2. Solution Dear, The answer is .2..
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
ankitmobileshop235
d. Increases venous PO2. systemic venous blood is likely to be 25% saturated with oxygen. PO2 Partial presure of oxygen. Solution d. Increases venous PO2. systemic venous blood is likely to be 25% saturated with oxygen. PO2 Partial presure of oxygen..
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
ankitmobileshop235
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
Considering the general solution of the Laplace Equation we getXn(.pdf
ankitmobileshop235
Cranial nerves which do not innervate the structures of the head but innervate the structures originating from branchial arches are the following Solution Cranial nerves which do not innervate the structures of the head but innervate the structures originating from branchial arches are the following.
Cranial nerves which do not innervate the structures of the head but.pdf
Cranial nerves which do not innervate the structures of the head but.pdf
ankitmobileshop235
Code: //Include libraries. import javax.swing.JOptionPane; //Define a class lPayroll public class lPayroll { //Main function public static void main( String args[] ) { //Initialize variables String lfirst = \"\"; String lLast = \"\"; String lssn = \"\"; String lsalaryString; double lsalary = 0; String lwageString; double lwage = 0; String lhoursString; int lhours = 0; String lsalesString; double lsales = 0; String lrateString; double lrate = 0; String lempTypeString = \"\"; int lempType = 0; int lcount; for(lcount = 0; lcount < 5; lcount++) { //Receive user input lempTypeString = JOptionPane.showInputDialog(null, \"Enter employee type \ \ Enter your choice.\ 1. Salaried\ 2. Hourly\ 3. Commissioned\ 4. Commission plus base\", \"Employee Type\", JOptionPane.QUESTION_MESSAGE); lempType = Integer.parseInt(lempTypeString); switch(lempType) { case 1: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name \", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lsalaryString = JOptionPane.showInputDialog(null, \"Enter weekly salary\", \"lsalary\", JOptionPane.QUESTION_MESSAGE); lsalary = Double.parseDouble(lsalaryString); break; case 2: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lwageString = JOptionPane.showInputDialog(null, \"Enter hourly wage\", \"Hourly lwage\", JOptionPane.QUESTION_MESSAGE); lwage = Double.parseDouble(lwageString); lhoursString = JOptionPane.showInputDialog(null, \"Enter count of hours per week of work\", \"lhours Worked\", JOptionPane.QUESTION_MESSAGE); lhours = Integer.parseInt(lhoursString); break; case 3: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_MESSAGE); lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\", JOptionPane.QUESTION_MESSAGE); lsalesString = JOptionPane.showInputDialog(null, \"Enter gross sales on weekly basis\", \"Gross lsales\", JOptionPane.QUESTION_MESSAGE); lsales = Double.parseDouble(lsalesString); lrateString = JOptionPane.showInputDialog(null, \"Enter commission percent\", \"Commsission lrate\", JOptionPane.QUESTION_MESSAGE); lrate = Double.parseDouble(lrateString); break; case 4: lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\", JOptionPane.QUESTION_MESSAGE); lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\", JOptionPane.QUESTION_ME.
Code Include libraries. import javax.swing.JOptionPane;.pdf
Code Include libraries. import javax.swing.JOptionPane;.pdf
ankitmobileshop235
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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