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c. A destructor to free any allocated objects. d. A copy constructor that allocates its own new member storage and copies the contents of member variables. e. An assignment operator that deallocates old storage before allocating new storage and copying all the member variables. Solution c. A destructor to free any allocated objects. d. A copy constructor that allocates its own new member storage and copies the contents of member variables. e. An assignment operator that deallocates old storage before allocating new storage and copying all the member variables..

c. A destructor to free any allocated objects.d. A copy constructo.pdf

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Barrier to entry: The government intervention to the barriers to entry to exist as a result of, while others occur naturally within the business world. Often, existing firms within an industry lobby for the government to erect new barriers to entry. In this is done to done by the protect the integrity of the industry and prevent fly-by-night operations from setting up shop and hawking inferior products and services. In reality, firms favor barriers to entry when already comfortably ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue. Other barriers to entry occur naturally, often evolving over time as certain industry players establish dominance. A company can use this technology, for example, to build a barrier to entry, to build in switching costs, and even, sometimes, to completely change the basis of competition. some companies have seized the advantage, while others, more complacent, have ended up playing the difficult and expensive game of catch-up ball. He also points out that it is important for executives to make this competitive analysis in assessing where IS fits in their companies, since in some cases it appropriately plays a support role and can add only modestly to the value of a company’s products, while in other settings it is at the core of their competitive survival. The computer’s main purpose is to cut order-entry costs and to provide more flexibility to customers in the time and process of order submission. The system yields a larger competitive advantage, adding value for customers and a substantial rise in their sales. The resulting sharp increase in the company’s market share forces a primary competitor into a corporate reorganization and a massive systems development effort to contain the damage. Natural Barriers to Entry: In industries where customers incur high costs switching from one brand to another, this becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective customers to pay the money required to make a chahange. Brand identity and customer loyalty serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics of an industry take shape of the natural barrier. Barrier option: A barrier option is a type of option whose payoff depends on whether or not the underlying asset has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but limiting losses for the writer. It can also be a knock-in, meaning it has no value until the underlying reaches a certain price. BREAKING DOWN \'Barrier Option\': Barrier options are considered a type of exotic option because they are more complex than basic American or European options. Barrier options are also considered a type of path-dependent option because their value fluctuates as the underlying\'s value changes during the opt.

Barrier to entryThe government intervention to the barriers to en.pdf

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Answer D all of the aboveLipids like triglycerides are semi-solid.pdf

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An example of when a one-way ANOVA could be used is if you want to determine if there is a difference in the mean height of stalks of three different types of seeds. Since there is more than one mean, you can use a one-way ANOVA since there is only one factor that could be making the heights different. Now, if take these three different types of seeds, and then add the possibility that four different types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of the stalks could be different for a combination of several reasons: The types of seed could cause the change, the types of fertilizer could cause the change, and/or there is an interaction between the type of seed and the type of fertilizer. There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then you can use a two-way ANOVA. Solution An example of when a one-way ANOVA could be used is if you want to determine if there is a difference in the mean height of stalks of three different types of seeds. Since there is more than one mean, you can use a one-way ANOVA since there is only one factor that could be making the heights different. Now, if take these three different types of seeds, and then add the possibility that four different types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of the stalks could be different for a combination of several reasons: The types of seed could cause the change, the types of fertilizer could cause the change, and/or there is an interaction between the type of seed and the type of fertilizer. There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then you can use a two-way ANOVA..

An example of when a one-way ANOVA could be used is if you want to d.pdf

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10.dry and hot conditions 11.C4 plants 12.bundle sheet cells 14.plasmodesmata 15)PEP carboxylase 16) bundle sheet cells 17) pineapple 18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if takes place carboxylase activity will involve 19)corn Solution 10.dry and hot conditions 11.C4 plants 12.bundle sheet cells 14.plasmodesmata 15)PEP carboxylase 16) bundle sheet cells 17) pineapple 18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if takes place carboxylase activity will involve 19)corn.

10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf

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1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf

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1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic alkalosis which is causing high HCO3 values. 2) the pH is alkalotic and it is abnormal due to metabolic alkalosis Solution 1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic alkalosis which is causing high HCO3 values. 2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.

1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf

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The intermediate is the compound A. There are a couple of reasons for that: 1. The oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic. Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom (the one that bears OH and then gets into the cycle) does not change in the reaction (retention of the configuration). That supports the formation of the intermediate A. If the intermediate B participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion of the configuration). Solution The intermediate is the compound A. There are a couple of reasons for that: 1. The oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic. Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom (the one that bears OH and then gets into the cycle) does not change in the reaction (retention of the configuration). That supports the formation of the intermediate A. If the intermediate B participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion of the configuration)..

The intermediate is the compound A. There are a c.pdf

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When ionic solids dissolve in water theions that are adjacent to each other in the solid become surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to large ions. If this formation of the hydratedions were the only factor than we would expect all ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we need to condsider the other factors in thesolution process. the solution process is governed bythe solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent interactions. When we consider the soluteinteractions we begin to see some of the problems that can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a solidis a measure of the strength of those electrostatic attractions.The lattice energy works to keep the ions in the solid state andionic compounds with large lattice energies are insoluble in waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g) ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation; H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3 is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms as the energy required to separate the ions in thesolute and the heat of solution isendothermic. Solution When ionic solids dissolve in water theions that are adjacent to each other in the solid become surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to large ions. If this formation of the hydratedions were the only factor than we would expect all ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we need to condsider the other factors in.

When ionic solids dissolve in water theions that are adjacent to each.pdf

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#include #include #include #include #include using namespace std; //Used from int index_of_largest(const int array[], int startIndex, const int size); void selectSort(int a[], const int size); void swap(int& i1, int& it2); void fileOption(); void inputOption(); void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size); void printFileScreen(const int uniqueArray[], const int countArray[], const int size); int main() { // int size; char userInput[5]; // cout << \"Read file from Input file? (Y/N) \"; cin >> userInput; if(toupper(userInput[0]) == \'Y\') { fileOption(); } else { inputOption(); } return 0; } void selectSort(int array[], const int size) { int smallestIndex; for(int i = 0;i < size - 1;i++) { smallestIndex = index_of_largest(array , i , size); swap(array[i] , array[smallestIndex]); } } void swap(int& i1, int& i2) { int temp; temp = i1; i1 = i2; i2 = temp; } int index_of_largest(const int array[], int startIndex, const int size) { int min = array[startIndex] , index_of_min = startIndex; for(int i = startIndex + 1;i < size;i++) if(array[i] > min) { min = array[i]; index_of_min = i; } return index_of_min; } void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size) { int number; int j = 0; int index; for(int i = 0;i < size;i++) { int flag = 0; number = inputArray[i]; //for each element in inputArray, check if it is there in uniqueArray for (int k = 0; k < j; ++k) { if(number == uniqueArray[k]) { flag = 1; index = k; } } if(flag == 0) { //if so, increment the countArray by 1 countArray[j]++; // copy the element to uniqueArray uniqueArray[j] = number; j++; } else countArray[index]++; } } void fileOption() { int inputArray[50]; int countArray[50] = {0}; int uniqueArray[50]= {0}; ifstream in_stream; char fileName[16]; cout << \"What file would you like to obtain input from? \"; cin >> fileName; in_stream.open(fileName); if(in_stream.fail()) { cout << \"Error: Could not open file!\" << endl; exit(1); } int size = 0; while(in_stream >> inputArray[size]) { size++; } selectSort(inputArray , size); unique(inputArray , uniqueArray , countArray , size); printFileScreen(uniqueArray , countArray , size); } void inputOption() { int inputArray[50]; int countArray[50] = {0}; int uniqueArray[50] = {0}; int size; cout << \"How many numbers would you like to enter? \"; cin >> size; cout << \"Enter numbers:\" << endl; for(int i = 0;i < size;i++) { cin >> inputArray[i]; } selectSort(inputArray , size); unique(inputArray , uniqueArray , countArray , size); printFileScreen(uniqueArray , countArray , size); } void printFileScreen(const int uniqueArray[], const int countArray[], const int size) { //cout << \"jaja\"; ofstream output_file; output_file.open(\"lab3Exercise2.txt\"); if(output_file.fail()) { cout << \"Error: Could not open file!\" << endl; exit(1); } output_file << \"N\\t\" << \"count\ \"; for(int i = 0;i < size;i++) { if(uniqueArray[i] != 0) { output_file << uniqueArray[i] << \"\\t\"; outpu.

#include iostream #include fstream #include map #include.pdf

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r=(13)d(I-)dt .pdf

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You have the priority for 1 right (the OH), however the positions 2 and 3 as you have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group, however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority 2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H is projected outwards so you get the opposite configuration of what you see (if it looks like an S you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of the molecule, therefore the R is really an S. Hope this helped! Solution You have the priority for 1 right (the OH), however the positions 2 and 3 as you have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group, however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority 2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H is projected outwards so you get the opposite configuration of what you see (if it looks like an S you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of the molecule, therefore the R is really an S. Hope this helped!.

You have the priority for 1 right (the OH), howev.pdf

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a electrolytic solution because it is an ionic co.pdf

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conjugated dienes are those in which two double b.pdf

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Fluorine is the most reactive and most electronegative of the elements. Fluorine also has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner. As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the other hand is considerably less electronegative than fluorine and it\'s electron affinity is considerably less exothermic than that of fluorine. Solution Fluorine is the most reactive and most electronegative of the elements. Fluorine also has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner. As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the other hand is considerably less electronegative than fluorine and it\'s electron affinity is considerably less exothermic than that of fluorine..

Fluorine is the most reactive and most electroneg.pdf

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hey dear neutron and protons do not occupy orbita.pdf

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at 100 degree KOH needs less Ea which cannot be p.pdf

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