Questions1) How does multimedia transmission differ than messag.pdf

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Questions: 1) How does multimedia transmission differ than \'message\' transmission? 2) Describe compression and include the different types used in multimedia. Solution 1) Message transmission includes the standard text communication using standard communication protocols. Here the messages are generally short and fixed in length. Message Service Centres (MSCs) are responsible for sending and receiving of messages. These messages are stored and forwarded and in case of any failure, they are queued for resending. The message if successfully sent, is then discarded from storage. Multimedia transmission is a way to send messages that includes multimedia content like images, music or video files. Message Service Centres along with carriers are responsible for successful sending and receiving of messages. Carriers are mainly used to change the format of the Multimedia message because the formats supported on different targeted devices are different. Internet is required to transfer multimedia messages between carrier and carrier. If the target device does not support multimedia, these messages can be viewed on the browser through an URL send as text message. 2) Compression is reduction in size of a file by reducing the number of bits that are used to represent without any loss of data. Generally, compression is employed to save storage space and increase the pace of file sending over a network which in turn saves bandwidth. An algorithm is used to compress a file. Multimedia compression techniques are classified as Lossy Compression: These recover the somewhat a similar representation to the original file. They have high compression ratios and used more practically. Examples of Lossy compression techniques are Lossless Compression: These techniques recover the original file. Examples of lossless techniques are.

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Fill in the blanks. Intensity of orange/brown/red color of reduced DNS is ____________ related to the amount of _______________ in solution. Solution Intensity of orange/brown/red color of reduced DNS is proportionally related to the amount of maltose in solution. DNS is used to study the enzymatic activity of amylase, which breaks starch into maltose. Higher amylase activity produces more amount of maltose, which reduces 3, 5-dinitrosalicylic acid (DNS), which is yellow in color, to 3-amino, 5-nitrosalicylic acid (commonly termed \"reduced DNS\"), which may be orange, brown or red depending on the concentration of maltose..

Fill in the blanks.Intensity of orangebrownred color of reduced .pdf

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Explain how cellular respiration (O2 consumption and CO2 production) is linked to ATP production. Is the coupling of these two processes perfect? Why or why not? Solution The primary function of cellular respiration is to generate ATP, which traps some of the chemical energy of food molecules in its high- energy bonds (adenosine triphosphate). The process of generation of ATP is via glycolysis and Krebs’s cycle finally through oxidative phosphorylation. The overall balanced reaction of cellular respiration is: CHO + 6O 6CO + 6HO + ATP Glucose + oxygen carbon dioxide + water + energy In this reaction, glucose oxidized and oxygen reduced. Cellular respiration (O2 consumption and CO2 production) is linked to ATP production because it is different from cellular respiration as it involves metabolic activities using molecular oxygen such as Citric acid cycle and electron transport to synthesize energy in the form of ATP to perform respiration. CO2 is the output of the cellular respiration from mitochondria and expelled out of the body by carrying through hemoglobin in the form of carboxy- hemoglobin to the alveoli of lungs. The coupling of these two processes such as O2 consumption and CO2 production inked to ATP production is perfect. Mitochondrial chemiosmosis –cellular respiration (through Fernadez Moran inner mitochondrial oxysomes particles) during cellular respiration obtain energy from chemical enzymatic breakdown of the organic food molecules (glucose, pyruvate, acetylcoA) to produce ATP in the presence of O2 consumption. During electron transport, the final electron acceptor is oxygen for oxidative phosphorylation via energy producing molecules such as NADH (3 ATP), FADH2 (2ATP) and ATP. Therefore, it is going to \"oxygen\" that accepting electrons at the end of the electron transport chain to generate ATP via mitochondrial respiratory complex- I to V. Every 3 protons used to produce one ATP molecule. Inner membrane possess small protein channels known as porins in mitochondria & these channels promote the movement of any small molecules such as ATP through them However, the coupling is perfect between the O2 consumption and CO2 production for ATP production but whole-organism oxygen consumption is mainly related to rate of substrate oxidation & it cannot be accurate measurement for ATP production due to variable P/O ratio occurs..

Explain how cellular respiration (O2 consumption and CO2 production).pdf

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Entrepreneurship is A. one of the basic rights of the private enterprise system. B. the willingness to take risks. C. required for capital expansion. D. not found in the private enterprise system. Solution B. the willingness to take risks. entrepenuership is the willingness to take risk. even if u have money and idea of business, but u don\'t have willpower to invest in that idea and take risk than u can\'t be entrepreneure. so risk taker can only become entrepenure than only he can set up a business by using his own or borrowed capital..

Entrepreneurship isA. one of the basic rights of the private enter.pdf

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Discrete Random Variables we are working with commonly used discrete and continuous distribution In a hospital there are 16 patients, 4 of whom have AIDS. A doctor is assigned to 6 of these patients at random. What is the probability that he gets 2 of the AIDS patients? Solution There are 16C6 ways to choose any 6 patients. There are 4C2 ways to get 2 with AIDS, and 12C4 to choose 4 without. Thus, P = [(12C4)(4C2)]/(16C6) = 0.370879121 [ANSWER].

Discrete Random Variables we are working with commonly used discre.pdf

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Determine the open intervals over which the function is increasing an.pdf

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Discuss and define Isotropy. Anisotropy and Orthotropy with regards to engineering materials. Regarding Finite Element Analysis: List the three main types of elements and explain how to choose elements? Common sources of errors in FEA? After conducting a photoelastic investigation of a sample, you remove the load and notice that a fringe pattern still exists. What could this mean? Solution (i) Isotropy is the property of the material in which the physical properties of a substance are independent of the direction of measurement. Glass and metals are examples of isotropic materials. Anisotropy is the property of the material in which the physical properties of a substance varies with the change in direction. Wood and Composites are examples of anisotropic materials. Orthotropy is the property of the material in which the physical properties of a substance change when measured from different directions. The properties differ along three mutually orthogonal twofold axes of rotational symmetry. Wood is an example of orthotropic materials. (ii) (a )The three main elements in FEA are -.

Discuss and define Isotropy. Anisotropy and Orthotropy with regards t.pdf

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Consider the circuits in figure 2.32. If we assume the transistors are well described by the Schockley model and their respective S and B terminals are tied together, the two circuits are equivalent (i.e. I_dsi=I_ds2). If the body effect is considered, will I_Ds2 be equal to, greater than, or less than I_dsi? Consider only body effect: assume velocity saturation and mobility degradation effects are negligible. Explain. Solution As source(S) and Substrate terminals are tied together, hence VSB= potential difference between S and B=0. Hence is no body effect. Hence Id1= Id2..

Consider the circuits in figure 2.32. If we assume the transistors ar.pdf

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Can you please debug this? Thank you in advance! This program is supposed to use stacks and find the paths of an airplane. It should tell you whether or not the plane goes from the requested city to the requested destination. requestFile.txt contains the requests, flightFile.txt contains the flights and cityFile.txt contains the city\'s that the airline serves. My errors: my output is simply \"Hpair does not serve Los Angeles.\" Thats it and it stops. It should look something like (these are not real values of the file just an example): \"Request is to fly from cityA to CityB. Hpair flight from cityA to CityB. Request is to fly from cityC to cityD. Sorry, HPAir does not serve City D Request is to fly from cityE to CityF Sorry, HPAir does not fly from cityE to CityF\" **It said to use ispath.cpp but I got confused on how to implement it completely.\" isPath.cpp //***I did not include ispath.cpp in my program. Parts of it are in the map.h portion. main.cpp #include #include #include \"Map.h\" using namespace std; int main() { Map aMap; aMap.readFlights(); aMap.verifyRequestedFile(); } StackInterface.h #ifndef StackInterface_h #define StackInterface_h template class StackInterface { /** Checks if the stack is empty. @return True if the stack is empty or false if the stack is not.*/ virtual bool isEmpty() const = 0; /** Adds a new item to the stack @post newEntry is at the top of the stack @param newEntry is the object to be added to the stack @return True if successfully added, otherwise false. */ virtual bool push(const ItemType& newEntry) = 0; /** Removes item from stack @post The top item in the stack is removed @return True if the item was removed, false if it was not. */ virtual bool pop() = 0; /** Returns the top item in the stack @pre The stack is not empty @post The top item in the stack is returned @return The top of the stack */ virtual ItemType peek() const = 0; }; #endif /* StackInterface_h */ Node.h #ifndef node_h #define node_h template class Node { public: Node(); Node (const ItemType &anItem); Node(const ItemType &anItem, Node * nextNodePtr); void setItem(const ItemType &anItem); void setNext(Node * nextNodePtr); ItemType getItem() const; Node* getNext() const; private: ItemType item; Node* next; }; template Node::Node():next(nullptr){} template Node::Node(const ItemType &anItem):item(anItem), next(nullptr){} template Node::Node(const ItemType &anItem, Node * nextNodePtr):item(anItem), next(nextNodePtr) { } template void Node::setItem(const ItemType &anItem){ item = anItem; } template void Node::setNext(Node * nextNodePtr) { next = nextNodePtr; } template ItemType Node::getItem() const { return item; } template Node* Node::getNext()const{ return next; } #endif /* node_h */ LinkedStack.h #include #ifndef LinkedStack_h #define LinkedStack_h #include \"Node.h\" #include \"StackInterface.h\" template class LinkedStack:public StackInterface{ public: LinkedStack(); LinkedStack(const LinkedStack &aStack); //space matter?*********** virt.

Can you please debug this Thank you in advance! This program is sup.pdf

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