1. Mucoviscidosis2. Digestive system and lungs3. symptoms- mucus.pdf

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1. Mucoviscidosis 2. Digestive system and lungs 3. symptoms- mucus accumutes, poor weight gain, salty skin, chest infections. 4. physiological signs- diarrhoea, cough, pneumonia, pale stools. Anatomical- accumulation of mucus, less weight despite of normal diet. Solution 1. Mucoviscidosis 2. Digestive system and lungs 3. symptoms- mucus accumutes, poor weight gain, salty skin, chest infections. 4. physiological signs- diarrhoea, cough, pneumonia, pale stools. Anatomical- accumulation of mucus, less weight despite of normal diet..

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Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes. In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures specialized in mediating the bidirectional traffic of molecules between the nucleus and the cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009). Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is responsible for the protected exchange of components between the nucleus and cytoplasm and for preventing the transport of material not destined to cross the nuclear envelope(UIUC). Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo proteinmust associate with another protein called a transport receptor. These transport receptors then act as chaperones which shuttle cargo through the NPC. Recognition by the transport receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo and transport receptors, the third ingredient necessary for nuclear transport is the signaling protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the interaction of transport receptor and cargo and RanGDP/RanGTP concentration gradients across the nuclear envelope drive nuclear import and export. Once a transport receptor identifies and binds its cargo, the transport complex may be imported or exported through the NPC. Repeating amino acid sequences involving phenyalanine (F) and glycine (G) have been thoroughly implicated in the process selective import,(UIUC). Solution Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes. In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures specialized in mediating the bidirectional traffic of molecules between the nucleus and the cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009). Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is responsible for the protected exchange of components between the nucleus and cytoplasm and for preventing the transport of material not destined to cross the nuclear envelope(UIUC). Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo proteinmust associate with another protein called a transport receptor. These transport receptors then act as chaperones which shuttle cargo through the NPC. Recognition by the transport receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo and transport receptors, the third ingredient necessary for nuclear transport is the signaling protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the inter.

Thepossible mechanism by which a 75kD nuclear protein is imported in.pdf

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The neurons are nervous cells which play the central role in generating, holding and transmitting information across the body. The information as we perceive is a physico-physiological aspect of our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally understood as: The neurons comprise of a complex set of neurological membranes and ion-channels. When a stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of various electrically stimulated ions present inside the neurons and outside their perphery. This causes generation of a local ionic gradient in the neurons which further drives physical activity of the neuron. The neurological membranes play important role in relay of information. These membranes help in relay of these biochemical signals across various neurological membranes and thus transit this information from one neuron to another via a matrix between the two, called synapse. This process takes place in a very small fraction of seconds so that an immediate and appropriate reaction could be made. Thus, a neurological cell holds the information by means of generation of a local bio-chemical potential and ionic gradient and relays/moves this information by means of channelizing these local gradients in the form of electro-chemical signals from one neuron to the other through synapse. Solution The neurons are nervous cells which play the central role in generating, holding and transmitting information across the body. The information as we perceive is a physico-physiological aspect of our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally understood as: The neurons comprise of a complex set of neurological membranes and ion-channels. When a stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of various electrically stimulated ions present inside the neurons and outside their perphery. This causes generation of a local ionic gradient in the neurons which further drives physical activity of the neuron. The neurological membranes play important role in relay of information. These membranes help in relay of these biochemical signals across various neurological membranes and thus transit this information from one neuron to another via a matrix between the two, called synapse. This process takes place in a very small fraction of seconds so that an immediate and appropriate reaction could be made. Thus, a neurological cell holds the information by means of generation of a local bio-chemical potential and ionic gradient and relays/moves this information by means of channelizing these local gradients in the form of electro-chemical signals from one neuron to the other through synapse..

The neurons are nervous cells which play the central role in generat.pdf

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The parts of the respiratory tract which have nonkeratinized stratif.pdf

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The impact of Brexit would be visible on the financial instituations of US and Europe. The impact would be shorter and longer term both. The biggest of all is the uncertainty it has created. So at first the financial firms would hold on to the likely investments they are going to make in UK as they would go in wait and watch mode. Also they are likely going to wrap up the operations or shift their base from UK to other European country assuming that the base operations caters primarily to the European nations. This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses would turn attractive due to currency depreciation. Solution The impact of Brexit would be visible on the financial instituations of US and Europe. The impact would be shorter and longer term both. The biggest of all is the uncertainty it has created. So at first the financial firms would hold on to the likely investments they are going to make in UK as they would go in wait and watch mode. Also they are likely going to wrap up the operations or shift their base from UK to other European country assuming that the base operations caters primarily to the European nations. This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses would turn attractive due to currency depreciation..

The impact of Brexit would be visible on the financial instituations.pdf

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The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO, have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature. The enthalpies of formation of these oxides are quite high and consequently they are very stable to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides except BeO are basic in nature and react with water to form sparingly soluble hydroxides. Solution The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO, have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature. The enthalpies of formation of these oxides are quite high and consequently they are very stable to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides except BeO are basic in nature and react with water to form sparingly soluble hydroxides..

The alkaline earth metals burn in oxygen to form the monoxide, MO wh.pdf

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Squid: mollusc Crab: Arthropods Crayfish: Arthropods Polychaete: Annelids Homologous organs are the one that develop from common ancestor and have diverged to perform different functions according to the habitat and habit of the organisms. Lateral appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit homology. The lateral appendages (parapodia) in polychaete worms extend from the body segments and serve in locomotion. However, soft body plan does not fix them firmly. On the other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the muddy oceanic sediments. Solution Squid: mollusc Crab: Arthropods Crayfish: Arthropods Polychaete: Annelids Homologous organs are the one that develop from common ancestor and have diverged to perform different functions according to the habitat and habit of the organisms. Lateral appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit homology. The lateral appendages (parapodia) in polychaete worms extend from the body segments and serve in locomotion. However, soft body plan does not fix them firmly. On the other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the muddy oceanic sediments..

Squid molluscCrab ArthropodsCrayfish ArthropodsPolychaete .pdf

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The code you written is little bit confusing and lengthy. Here is the efficient code to your problem. import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.util.HashMap; import java.util.Map; import java.util.Scanner; public class ZooMonitor { private static Scanner scanner=new Scanner(System.in); public static void main(String[] args) { int choice; while(true){ System.out.println(\"------ Menu ------\"); System.out.println(\"1. Monitor an animal\"); System.out.println(\"2. Monitor a habitat\"); System.out.println(\"3. Exit\"); choice=scanner.nextInt(); switch(choice){ case 1: monitorAnimal(); break; case 2: monitorHabitat(); break; case 3: System.exit(0); default: System.out.println(\"Invalid Input\"); } } } public static void monitorAnimal(){ int choice; BufferedReader br = null; int count=0; Map animals=new HashMap(); System.out.println(\"------ Animals ------\"); try { String sCurrentLine; br = new BufferedReader(new FileReader(\"animals.txt\")); while ((sCurrentLine = br.readLine()) != null) { if(sCurrentLine.startsWith(\"Details\")){ String animal=sCurrentLine.split(\" \")[2]; System.out.println(++count+\". \"+animal); animals.put(count, animal); } } br.close(); br = new BufferedReader(new FileReader(\"animals.txt\")); System.out.println(\"Enter choice : \"); choice=scanner.nextInt(); String selectedAnimal=animals.get(choice); if(selectedAnimal!=null){ while ((sCurrentLine = br.readLine()) != null) { if(sCurrentLine.startsWith(\"Animal\")&&(sCurrentLine.split(\" \")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\" \")[2].toLowerCase()))){ for(int i=0;i<5;i++){ if(sCurrentLine.startsWith(\"*\")) System.err.println(sCurrentLine.substring(5)); else System.out.println(sCurrentLine); sCurrentLine=br.readLine(); } } } } } catch (IOException e) { e.printStackTrace(); } finally { try { if (br != null)br.close(); } catch (IOException ex) { ex.printStackTrace(); } } } public static void monitorHabitat(){ int choice; BufferedReader br = null; int count=0; Map habitats=new HashMap(); System.out.println(\"------ Habitats ------\"); try { String sCurrentLine; br = new BufferedReader(new FileReader(\"habitats.txt\")); while ((sCurrentLine = br.readLine()) != null) { if(sCurrentLine.startsWith(\"Details\")){ String habitat=sCurrentLine.split(\" \")[2]; System.out.println(++count+\". \"+habitat); habitats.put(count, habitat); } } br.close(); br = new BufferedReader(new FileReader(\"habitats.txt\")); System.out.println(\"Enter choice : \"); choice=scanner.nextInt(); String selectedAnimal=habitats.get(choice); if(selectedAnimal!=null){ while ((sCurrentLine = br.readLine()) != null) { if(sCurrentLine.startsWith(\"Habitat\")&&(sCurrentLine.split(\" \")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\" \")[2].toLowerCase()))){ for(int i=0;i<4;i++){ if(sCurrentLine.startsWith(\"*\")) System.err.println(sCurrentLine.substring(5)); else System.out..

The code you written is little bit confusing and lengthy. Here is th.pdf

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Ques-1: Normala flora and microbiota of nose and mouth are non-infectious in the region where they are living in colonies at specific pH and temperature. However, if they reach other parts of human body other than nose and mouth may induce “infection” They do possess common characteristics of sensing known as “quorum sensing” in which bacteria flora are going to communicate with each other in the region or habitat where they are living. They have a specific property of “horizontal gene transfer” Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and spirochetes are going to present in nose due to presence of moist environments Ques-2: Answer: Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota. There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of infection present in oral & nasal cavities Solution Ques-1: Normala flora and microbiota of nose and mouth are non-infectious in the region where they are living in colonies at specific pH and temperature. However, if they reach other parts of human body other than nose and mouth may induce “infection” They do possess common characteristics of sensing known as “quorum sensing” in which bacteria flora are going to communicate with each other in the region or habitat where they are living. They have a specific property of “horizontal gene transfer” Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and spirochetes are going to present in nose due to presence of moist environments Ques-2: Answer: Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota. There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of infection present in oral & nasal cavities.

Ques-1Normala flora and microbiota of nose and mouth are non-infe.pdf

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Solution.According to IASB framework INCOME is the one of element .pdf

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7. Answer is A. Cytochromes are protien containing iron or heme groups and function as electron transfer agents in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t has sulfur groups and it doesn\'t reduce oxygen to water. 8. Answer is B. Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH. 9. Answer is E. The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has potential difference of 1.136 V; the electron flow will be from the most negative system, which is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless enzymes are present to catalyze the reaction. Solution 7. Answer is A. Cytochromes are protien containing iron or heme groups and function as electron transfer agents in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t has sulfur groups and it doesn\'t reduce oxygen to water. 8. Answer is B. Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH. 9. Answer is E. The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has potential difference of 1.136 V; the electron flow will be from the most negative system, which is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless enzymes are present to catalyze the reaction..

7. Answer is A.Cytochromes are protien containing iron or heme gro.pdf

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A and C. RGB analog component video uses no compression, imposes no real limit on color depth and resolution, requires a large amount of bandwidth to carry a signal, and is being superseded by DVI and HDMI, which provide better clarity. Solution A and C. RGB analog component video uses no compression, imposes no real limit on color depth and resolution, requires a large amount of bandwidth to carry a signal, and is being superseded by DVI and HDMI, which provide better clarity..

A and C. RGB analog component video uses no compression, imposes no .pdf

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Public key authentication is the most secure colution and utilizes a public key cryptography(PKI) system, using both public and private keys.A PKI is certainly the easiest solution from the users prespective.It mainly supports the distribution and identification of public encryption keys, enabling users and computers to both securely exchange data over networks such as the internet and verify the identity of the other party. Without PKI , sensitive information can still be encrypted ( ensuring confidentiality) and exchanged but there would be no assurance of the identity(authentication) of the other party. Any form of sensitive data exchanged over the internet is relianton PKI for security. Elements of PKI A typical PKI consists of hardware, software , policies and standards to manage the creation,admisinstration , distribution and revocation of keys and digital certificates. Digital certificates are at the heart of PKI as they affrim the identity of the certificate subject that bind the identity to the public key contained in the certificate. Following are the elements of the PKI: Certificate Authority (CA) Registration authority A certificate database , it mainly stores the certificate requests and issues and revokes certificates. A certificate store , which resides on a local computer as a place to store issues certificates and private keys. --------------------------************************----------------------------- To connect many distant employees at once, all office locations must be able to access the same network resource. Most time iof we will would house the core infrastucture needed to establish the network, such as the servers and databases. Generally Wide area network (WAN) to connect the remote offices. WAN are used to connect Local area network(LAN) for disparate offices together. However installation and overhead of LNS and WANS is expensive. To over come this Virtual private networks that run over the internet. VPN it mainly provides secure connections between individual users and their organizations network over and their organizations network over the internet, have several upsides. They provide more secure site to site connections. They provide information transfer much faster that WAN\'s and most importantly to small and medimum sized business, VPN\'s are much lessexpensive since you can use a single leased line to the internet for each ofice, cutting down on broadband cost. While VPN\'s are highly recommended and trusted a new technology gaining traction with compaines is the remote desktop. Remote dsktops requires software or an operating system that allows aplications to run remotely on a server, but to be displayed locally simultaneously. This network server hosts remote files and application in a secure location , ensuring your data is never lost , even when something happens to device you are working on. Solution Public key authentication is the most secure colution and utilizes a public key cryptography(PKI) sy.

Public key authentication is the most secure colution and utilizes a.pdf

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Particle Charging Mechanisms in ESP: Particles are charged by negative gas ions moving toward the collection plate by one of these two mechanisms: 1. Field charging 2. diffusion charging. In field charging, particles capture negatively charged gas ions as the ions move toward the grounded collection plate. Diffusion charging, as its name implies, depends on the random motion of the gas ions to charge particles. In field charging, as particles enter the electric field, they cause a local dislocation of the field. Negative gas ions traveling along the electric field lines collide with the suspended particles and impart a charge to them. The ions will continue to bombard a particle until the charge on that particle is sufficient to divert the electric lines away from it. This prevents new ions from colliding with the charged dust particle. When a particle no longer receives an ion charge, it is said to be saturated. Saturated charged particles then migrate to the collection electrode and are collected. Diffusion charging is associated with the random Brownian motion of the negative gas ions. The random motion is related to the velocity of the gas ions due to thermal effects: the higher the temperature, the more movement. Negative gas ions collide with the particles because of their random thermal motion and impart a charge on the particles. Because the particles are very small, they do not cause the electric field to be dislocated, as in field charging. Thus, diffusion charging is the only mechanism by which these very small particles become charged. The charged particles then migrate to the collection electrode. Each of these two charging mechanisms occurs to some extent, with one dominating depending on particle size. Field charging dominates for particles with a diameter >1.0 micrometer because particles must be large enough to capture gas ions. Diffusion charging dominates for particles with a diameter less than 0.1 micrometer. A combination of these two charging mechanisms occurs for particles ranging between 0.2 and 1.0 micrometer in diameter. Solution Particle Charging Mechanisms in ESP: Particles are charged by negative gas ions moving toward the collection plate by one of these two mechanisms: 1. Field charging 2. diffusion charging. In field charging, particles capture negatively charged gas ions as the ions move toward the grounded collection plate. Diffusion charging, as its name implies, depends on the random motion of the gas ions to charge particles. In field charging, as particles enter the electric field, they cause a local dislocation of the field. Negative gas ions traveling along the electric field lines collide with the suspended particles and impart a charge to them. The ions will continue to bombard a particle until the charge on that particle is sufficient to divert the electric lines away from it. This prevents new ions from colliding with the charged dust particle. When a particle no longer receives an ion charge, it is.

Particle Charging Mechanisms in ESPParticles are charged by negat.pdf

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Option C Explanation: The word \'LEADING\' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. Solution Option C Explanation: The word \'LEADING\' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720..

Option CExplanationThe word LEADING has 7 diffe.pdf

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Please follow the data : There are some flag and the control characters that are usewd in the fields of communication to maintain a synchronous data from the sender and the receiver. Some of the flags described are the ACK and NAK flags or the control characters. These characters will generally have the standard decimal values that maintain. a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that has been initiated from the user or the sender. It also describes that the process was successful. It is the abbrevation of Acknowledgement. The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to binary we get the values of the ACK flag as (0110)2. Now the hexadecimal value can be given as the same as the values for the hexa and the binary would be same till the value of 15 as though they describr the same point. So the hexadecimal value of the ACK flag is (6)16. b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the data has not been received correctly or the data has been tampered or lost over the network. It is the abbrevation of Negative Acknowledgement. The decimal value that is listed accordingly for the characters is given by (21)10. So the binary format of the NAK flag is givem by (0001 0101)2. Now that converting it to the hexadecimal value we get the NAK flag value as 15(16). Hope this is helpful. Solution Please follow the data : There are some flag and the control characters that are usewd in the fields of communication to maintain a synchronous data from the sender and the receiver. Some of the flags described are the ACK and NAK flags or the control characters. These characters will generally have the standard decimal values that maintain. a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that has been initiated from the user or the sender. It also describes that the process was successful. It is the abbrevation of Acknowledgement. The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to binary we get the values of the ACK flag as (0110)2. Now the hexadecimal value can be given as the same as the values for the hexa and the binary would be same till the value of 15 as though they describr the same point. So the hexadecimal value of the ACK flag is (6)16. b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the data has not been received correctly or the data has been tampered or lost over the network. It is the abbrevation of Negative Acknowledgement. The decimal value that is listed accordingly for the characters is given by (21)10. So the binary format of the NAK flag is givem by (0001 0101)2. Now that converting it to the hexadecimal value we get the NAK flag value as 15(16). Hope this is helpful..

Please follow the data There are some flag and the control charac.pdf

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only anhydorus LiAlH4 acts as a good reducing agent ,that is i.pdf

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Part 1) #include #include #include #include /*Error handling for pthread_create and pthread_join*/ /*from the pthread_create man page*/ #define handle_error_en(en, msg) \\ do { errno = en; perror(msg); exit(EXIT_FAILURE); } while (0) /* # of running threads */ volatile int running_threads = 0; pthread_t thread[3]; /*Descriptors for our 3 threads*/ int numOfElements;/*Total # of elements from the user*/ struct Results{ /*Struct to hold the statistical results*/ int min; int max; int average; }Results; /*This function finds the minimum element of an array*/ void *findMin(void *array_ptr){ int i; /*counter*/ int *elements = (int*)array_ptr; /*re reference void array pointer*/ Results.min = elements[0]; /*set minimum to first element */ for(i = 0; i < numOfElements; i++){ /*iterate through array*/ if(elements[i] < Results.min){ /*if the current element is less than the current min*/ Results.min = elements[i]; /*store the new min*/ } } running_threads -= 1; /*Decrement thread count*/ return NULL; } /*This function finds the maximum element of an array*/ void *findMax(void *array_ptr){ int i; /*counter*/ int *elements = (int*)array_ptr; /*re reference void array pointer*/ for(i = 0; i < numOfElements; i++){ /*iterate through array*/ if(elements[i] > Results.max){ /*store the new max*/ Results.max = elements[i]; } } running_threads -= 1; /*Decrement thread count*/ return NULL; } /*This function finds the average of an array*/ void *findAverage(void *array_ptr){ int i; /*counter*/ int *elements = (int*)array_ptr; /*re reference void array pointer*/ for(i = 0; i < numOfElements; i++){ /*iterate through array*/ Results.average += elements[i]; /*add element @ i to average*/ } Results.average = Results.average/numOfElements; /*Divide the sum by the number of elements*/ running_threads -= 1; /*Decrement running threads counter*/ return NULL; } /* This method accepts a int n(initial size of array) and pointer to an array and returns # of elements in the array */ int getArrayInput(int n, int *array_ptr){ int input;/*Store user input */ int numberOfElements = 0;/*Number of Integers inputed*/ printf(\"Creating Dynamic Array...\ -\ \"); for(;;){ /*infinite loop*/ printf(\"Enter a positive value:\ Negative Number to Stop\ -\ \"); //Get Int from console, store at address of input if (scanf(\"%d\",&input) != 1){ printf(\"\ Oops that wasn\'t an Integer\ lets try filling the array again\ Remember INTEGERS only!\ \"); exit(EXIT_FAILURE); } if (input >= 0){ if (numberOfElements == n){ n += 1; //Make room for the current input array_ptr = realloc(array_ptr, n * sizeof(int));//realloc array and set pointer } array_ptr[numberOfElements++] = input;//Store input at next empty element } else { printf(\"\ Number of Integers: %d\ \", numberOfElements); break; } } return numberOfElements; } /*This function joins our n number of threads */ void joinThreads(int numberOfThreads){ int i; /*count*/ int s; /*error #*/ while(numberOfThreads >= 0){ /*Join our threads*/ s = pthread_jo.

Part 1)#include stdio.h #include stdlib.h #include pthrea.pdf

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Note that there are two poles in the system. The common region of the entire system is inside the one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the signal is left-sided. If it is greater than unit circuit, it is a right-sided signal. If it is between the poles, it is two-sided. Solution Note that there are two poles in the system. The common region of the entire system is inside the one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the signal is left-sided. If it is greater than unit circuit, it is a right-sided signal. If it is between the poles, it is two-sided..

Note that there are two poles in the system. The common region of th.pdf

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Obvious first to third shell. E is proportional .pdf

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a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent. Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful Solution a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent. Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful.

a) YES. Both are hydrocarbons and nonpolar. Like .pdf

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