Question 1 of 10 Map sapling learning Based on the chart to the right.pdf

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Question 1 of 10 Map sapling learning Based on the chart to the right Ratio of Exports to GDP for Major Countries and the World (Percent) in which year did world globalization hit its lowest point 35 30 O 1890 O 1913 25 France O 1960 Germany 20 O 1970 Japan 15 LUK. O 1990 US 10 In 1913, what country attained the World highest amount of globalization? 5 O France 1890 1970 1913 1960 1990 O Japan O Germany United O Kingdom (UK) United States (US) Solution A yardstick for measuring globalization is Exports/GDP ratio. The higher this ratio, the higher the magnitude of globalization. (a) World globalization was lowest in 1960, when Export/GDP ratio was lowest at 2.5% (b) In 1913, UK had highest globalization because its Export/GDP ratio was highest at 29%..

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In order to be useful to managers, management accounting reports should possess all of the following cs EXCEPT a, be prepared to report information for any unit of the business to support decision making. b. be ed at any time management needs information d. provide objective measures of past operations and subjective estimates about future decisions Solution The correct choice is b Explanation : - Management accounting is not required to adhere to generally accepted accounting principle because it provides managerial information for internal purposes i.e. it is used by the members within the organization and not used by members outside the organizations such as regulators, creditors etc..

In order to be useful to managers, management accounting reports shou.pdf

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Im also confused on question 9 and 10 On which point(s) would Charles Darwin and Georges Button have likely agreed Select one: the idea that all of life shared a common ancestor the idea that species change over time both are correct neither is correct On which point(s) would Charles Darwin and Georges Button have likely disagreed: Select one: the idea that the earth was formed according to the laws of physics the idea that all life shares a common ancestor the idea that species change over time they would have agreed on all of these points Which of the following ideas was NOT accepted by Jean-Batiste Lamarck? Select one: bacteria represent some of the youngest species on earth life arises spontaneously organisms go extinct none of the above Solution Answer c : Both are correct Answer d: Earth is quite old both agree, common ancestor in species tree Answer c:Lamarck did not believe that species could become extinct.

Im also confused on question 9 and 10 On which point(s) would Charle.pdf

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In about six billion years, our sun\'s luminosity (a.k.a. power output) will be 2000 times more than it is now. At this stage, our sun will be considered a \"red giant\" star. The surface temperature of a red giant is only about two thirds of our sun\'s current temperature. Between what two famous celestials objects will be the location of our sun\'s surface in six billion years? Solution Luminosity,L = k*R^2 * T^4 so, L2/L1 = [R2/R1]^2 * [T2/T1]^4 2000 = [R2/R1]^2 * 0.1975 R2 = 100.62 * R1 = 100.62 * 6.96*10^8 = 7.003 * 10 ^10m Radius of orbit of mercury = 4.6*10^10 m - 7 * 10 ^ 7m Radius of orbit of venus = 1.08*10^11 m Hence sun\'s surface will be between venus amd mercury.

In about six billion years, our suns luminosity (a.k.a. power outpu.pdf

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I have to write a java program . So basically I have to take my input in only cms. Inches is also accepted but the inches should be converted into the cms before the program take in the input . So when the program takes the input , the input must be in cm . I have to write a java program . So basically I have to take my input in only cms. Inches is also accepted but the inches should be converted into the cms before the program take in the input . So when the program takes the input , the input must be in cm . Solution import java.io.*; import java.util.*; class InchesToCms{ public static void main (String[] args)throws Exception{ //1 Inches = 2.54 Centimeters; BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));//to get input creating bf is an object. System.out.print(\"Enter the inches: \");//asking the input is inches float inches = Float.parseFloat(bf.readLine());//geting input first in string,so convert into float using parseFloat function. float cms = inches * 2.54f;//multiplying given input inches and 2.54. System.out.println(inches + \" inches is \" + cms + \" centimeter(s).\");//display the inches and centimeters. System.out.print(\"Enter the Centimeters: \");//asking the input is Centimeters. float CMS = Float.parseFloat(bf.readLine()); System.out.println(\"You entered \" +CMS+ \" centimeter(s).\");//display out put in Centimeters only. } } /* Output: 1) Enter the inches: 11.0 11.0 inches is 27.939999 centimeter(s). Enter the Centimeters: 2.0 You entered 2.0 centimeter(s). Explanatin: This program 1) It takes input in inches and convert into centimeters 2) and Next It takes input in centimters and display the out in the same centimeters. */.

I have to write a java program . So basically I have to take my in.pdf

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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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Circulating blood cells that differentiate into macrophages in the tissue M cells Monocytes Common myeloid progenitors Eosinophils Solution Answer: b. Monocytes Monocytes are the circulating blood cells which after production from the bone marrow, circulate in the bloodstream for about 8 h, during which they enlarge; and then migrate into the tissues and differentiate into specific tissue macrophages. The differentiation of a monocyte into a tissue macrophage involves a number of changes such as :.

Circulating blood cells that differentiate into macrophages in the ti.pdf

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Between January 2005 and May 2007, the CDC had identified 138 cases of Acanthomelobe Keratitis infections (a rare but potentially blinding infection of the cornea of the eye) in 35 states and Puerto Rico. Although only 46 interviews of infected individuals have been acquired, the results indicate a linkage to the use of Complete MoisturePlus Multipurpose contact lens solution. How might public health officials have handled this outbreak? Official letters to all registered healthcare practitioners announcing methods of treatment and follow-up Provide the option to health care providers to report the disease Public health announcement All of the abovea. Official letters to all registered healthcare practitioners announcing methods of treatment and follow-upb. Provide the option to health care providers to report the diseasec. Public health announcementd. All of the above Solution D.All of above Public health officials work in outbreak very precociously. Too often an initial suspicious outbreak is just lots of different illnesses that are being confused for the same thing or an error from the lab or other monitoring systems. So it’s important to study sign/symptoms and study the number of cases. The case definition will create a standardized method for identifying past, present and future cases. It should include clinical information, characteristics of those with the disease, and geographic and timing information about the cluster of cases. A good case definition starts out very inclusive so as not to miss any potential cases that do not have the typically identified symptoms. As the investigation continues for an unknown disease, this definition may become more restricted to ensure we focus on persons who truly have the illness of concern. After confirmation, they take appropriate action.As in above cases in only 46 interviews have been developed infection still part of their procedure they have to follow all three steps which mention above..

Between January 2005 and May 2007, the CDC had identified 138 cases .pdf

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A satellite is 112 miles above the earth, and the angle A to the hor.pdf

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A car antenna is 1.5 m long. The car is moving at (40 mph) under a DC high-tension wire with (100 amp) (the distance between the wire and the top of the antenna is 10m) . What emf is induced in the antenna and what is its polarity ? Solution emf is induced in the antenna = VBL 40mph = 17.881 m/s , B = u0I/(2piR) => emf = 17.881*4*3.14*10^-7100*1.5)/(2*3.14*10) = 0.000053643 V = 53.64 uV polarity will be in (VxB) direction.

A car antenna is 1.5 m long. The car is moving at (40 mph) under a D.pdf

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8:32 AM ut.blackboard.com ? HW #1-The Federal Reserve and Monetary Policy This homework assignment is worth up to 20 points toward your final grade. 1. What are the five (5) risks common to all financial institutions? Briefly explain each. (5 points) 2. Who is the current Chairman of the Federal Reserve? (2 points) 3. Explain the three uses of money. How do cryptocurrencies, such as Bitcoin and Ripple, fit into this framework? (3 points) 4. What are the three main goals of the Federal Reserve and how does the Fed use Monetary Policy to achieve these goals? (3 points) Solution Credit risk According to the Bank for International Settlements (BIS), credit risk is defined as the potential that a bank borrower or counterparty will fail to meet its obligations in accordance with agreed terms. Credit risk is most likely caused by loans, acceptances, interbank transactions, trade financing, foreign exchange transactions, financial futures, swaps, bonds, equities, options, and in the extension of commitments and guarantees, and the settlement of transactions. In simple words, if person A borrows loan from a bank and is not able to repay the loan because of inadequate income, loss in business, death, unwillingness or any other reasons, the bank faces credit risk. Similarly, if you do not pay your credit card bill, the bank faces a credit risk. Hence, to minimize the credit risk on the bank’s end, the rate of interest will be higher for borrowers if they are associated with high credit risk. Factors like unsteady income, low credit score, employment type, collateral assets and others determine the credit risk associated with a borrower. As stated earlier, credit risk can be associated with interbank transactions, foreign transactions and other types of transactions happening outside the bank. If the transaction at one end is successful but unsuccessful at the other end, loss occurs. If the transaction at one end is settled but there are delays in settlement at the other end, there might be lost investment opportunities. Look at it like person A sending US dollars to his family in India at the rate of 60 INR (Indian Rupee) per dollar. The person B, who is the recipient however receives the payment late and doesn’t get the exchange rate of 60 INR. Instead he receives the money at the exchange rate of 58 INR. This means they incurred a loss in the transaction. Similar situations occur during big transactions in banks. If the bank is not able to settle a transaction at an expected time or during an expected time duration, they may incur a credit risk. However, this kind of risk is called “Settlement Risk” and it is closely associated with credit risk. It depends on the timing of the exchange of value, payment/settlement finality and the role of intermediaries and clearing houses. While some credit risk is a result of macro forces affecting the economy or specific markets or even specific individuals, there is another important risk that can be classified un.

832 AM ut.blackboard.com HW #1-The Federal Reserve and Monetary Po.pdf

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you will use cellular automata to create a 2D predator–prey simulation in your program. The preys are ants and the predators are doodlebugs. These critters live in a 20 * 20 grid of cells. Only one critter may occupy a cell at a time. The grid is enclosed and no critter may move off the grid. Time is simulated in steps. Each critter performs some action every time step. The ants behave according to the following model: Move: For every time step, the ants randomly try to move up, down, left, or right. If the neighboring cell in the selected direction is occupied or would move the ant off the grid, then the ant stays in the current cell. Breed: If an ant survives for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left, or right) cell that is empty randomly. If there is no empty cell available, no breeding occurs. Once an offspring is produced, an ant cannot produce an offspring again until it has survived three more time steps. The doodlebugs behave according to the following model: Move: For every time step, the doodlebug will firstly try to move to an adjacent cell containing an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs. Breed: If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a new doodlebug in the same manner as the ant. Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time step it will starve and die. The doodlebug should then be removed from the grid of cells. Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them. You should see a cyclical pattern between the population of predators and prey, although random perturbations may lead to the elimination of one or both species. You will prompt the user to enter the number of time steps to run. Create a class named Critter that contains data and functions common to ants and doodlebugs. This class should have a virtual function named move that is defined in the derived classes of Ant and Doodlebug. Each class will be in its own source file. For each time step, do the following in your program: after moves, when breeding, eating, and starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O” for an ant, “X” for a doodlebug and “E” for an empty space (the characters should be arranged to look like a grid). The doodlebugs will move before the ants in each time step. When you reach the time steps entered by the user, ask them to enter another number and start to run the simulation again or to exit the program. You must maintain the state of the current grid while creating the next displa.

you will use cellular automata to create a 2D predator–prey simulati.pdf

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Why should you use properties within the methods of a class in C# even if syntactically you don’t have to ? Solution In general, methods represent actions and properties represent data. Properties are meant to be used like fields, meaning that properties should not be computationally complex or produce side effects. When it does not violate the following guidelines, consider using a property, rather than a method, because less experienced developers find properties easier to use.That way use properties..

Why should you use properties within the methods of a class in C# ev.pdf

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Which one is the answer? How do you know that? Occupancy type is a surrogate measure for Fire Load Ventilation Combustibility ToxicityA. Fire LoadB. VentilationC. CombustibilityD. Toxicity Solution Occupancy type is a surrogate measure for ventilation. Surrogate measures of human occupancy and activity in the 10 patient rooms using both indoor CO2 concentrations and infrared (IR) beam-break counters installed at the patient room doorways. Outdoor air fractions in the HVAC systems serving the two floors..

Which one is the answer How do you know thatOccupancy type is a .pdf

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