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Zoom in on your screen if needed. This is as small as a I can get the information. Please see the data in the next tab ("Data \& Template"). For this data, please implement a) Holt-Winter's forecasting method (please use the given values for alpha, beta and gamma - do *not* change them), and b) the four-period moving average forecasting method. Please answer the following questions: 1) What is forecast for period 51 (March of Year 5) when Holt-Winter's method is used? Answer: units 2) What is forecast for period 51 (March of Year 5) when the four-period moving average method is used? Answer: units Answer: \begin{tabular}{|c|l|r|r|} \hline 25 & January & 3 & 539 \\ \hline 26 & February & 3 & 620 \\ \hline 27 & March & 3 & 524 \\ \hline 28 & April & 3 & 665 \\ \hline 29 & May & 3 & 620 \\ \hline 30 & June & 3 & 563 \\ \hline 31 & July & 3 & 772 \\ \hline 32 & August & 3 & 720 \\ \hline 33 & September & 3 & 530 \\ \hline 34 & October & 3 & 772 \\ \hline 35 & November & 3 & 563 \\ \hline 36 & December & 3 & 641 \\ \hline 37 & January & 4 & 720 \\ \hline 38 & February & 4 & 577 \\ \hline 39 & March & 4 & 847 \\ \hline 40 & April & 4 & 865 \\ \hline 41 & May & 4 & 623 \\ \hline 42 & June & 4 & 806 \\ \hline 43 & July & 4 & 895 \\ \hline 44 & August & 4 & 691 \\ \hline 45 & September & 4 & 947 \\ \hline 46 & October & 4 & 970 \\ \hline 47 & November & 4 & 787 \\ \hline 48 & December & 4 & 1,024 \\ \hline 49 & January & 5 & 901 \\ \hline 50 & February & 5 & 1,028 \\ \hline 51 & March & 5 & 1,000 \\ \hline \end{tabular} .

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Write an SML function groupdupes that takes a list of integers as its only argument. The function takes runs of values (i.e., consectuive, duplicated values) in the list, copying those runs into separate sublists, returning a list of those sublists. Examples: groupdupes([4,4,4,2,3,3,7,7,7,7,2,3,4,4,4]) would return [[4,4,4],[2],[3,3],[7,7,7,7],[2],[3],[4,4,4]] groupdupes([8,6,7,5]) would return [[8],[6],[7],[5]] Solution fun groupdupes_helper(L:int list, subL:int list, result:int list list) = if (L = nil) then result @ [subL] else if ( ( subL = nil ) orelse ( hd(subL) = hd(L) ) ) then groupdupes_helper( tl(L), hd(L)::subL, result ) else groupdupes_helper( tl(L), hd(L)::nil, result @ [subL] ) (* groupdupes Wrapper Function *) fun groupdupes(L:int list) = groupdupes_helper(L, nil, nil); .

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You and another tech are discussing the relative merits of SCSI interfaces. You say that SCSI is dead, at least as far as drives go, and has been replaced by SAS and SATA. Your coworker says that SCSI has evolved and still has valid technical applications in todayâ€™s computing environment. Which modern applications does she say SCSI currently possesses? (Choose two.) A. SCSI now can use serial as well as parallel interfaces. B. SCSI now uses TCP/IP as a transport mechanism. C. Some motherboard manufacturers still market SCSI interfaces supporting PCI Express and PCI-X. D. SCSI Ultra-5 (parallel) now supports as 64-bit data width. Solution A. SCSI now can use serial as well as parallel interfaces. C. Some motherboard manufacturers still market SCSI interfaces supporting PCI Express and PCI-X. Although SCSI parallel (SPI) interfaces are still common, they are being replaced by Serial Attached SCSI (SAS), which retains the original SCSI technology but employs a serial design. Very few companies still market SCSI interfaces for motherboards supporting PCIe and PCI-X (PCI-X is a double-wide version of the PCI bus that satisfies the higher bandwidth requirements of servers). .

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Write an application class (ArrayListApplication) that contains a main(...) method. The method must perform the following. • Prompt user for 10 names and store in an \"ArrayList\" object • Shuffle the names using \"shuffle\"\" method of the \"Collections\" class • Display the smallest element in the collection use the \"Collections.min\" method. • Display the largest element in the collection use the \"Collections.max\" method. Solution import java.util.*; public class ArrayListNames { publicstaticvoid main(String[] args) { int i; ArrayList<String> name = new ArrayList<String>(); System.out.println(\"Enter 10 Names \"); for( i=0; i<10; i++ ) { Scanner s =newScanner(System.in); name add(s.next()); } System.out.println(\"Initial collection : \"+name); // shuffle the list Collections.shuffle(name); System.out.println(\"Final collection after shuffle: \"+name); Object minimum = Collections.min(name); System.out.println(\"Minimum Element of ArrayList name is : \" + minimum); Object maximum = Collections.max(name); System.out.println(\"Maximum Element of ArrayList name is : \" + maximum); } } .

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Write an algorithm for a program that shows the use of all six math functions. Write, test, and debug the program using Python. SAMPLE OUTPUT (not including author/program information) ADDITION: 2+2=4 SUBTRACTION: 4-2=2 MULTIPLICATION: 4*2=8 DIVISION: 4/2=2 EXPONENT: 2**3=8 REMAINDER: 5%2=1 Solution #!/user/bin/python def add(x, y): return x + y def subtract(x, y): return x - y def multiply(x, y): return x * y def divide(x, y): return x / y def exp(x, y): return x ** y def mod(x, y): return x % y print(\"Select any one operation what you want to perform.\") print(\"1.Add\") print(\"2.Subtract\") print(\"3.Multiply\") print(\"4.Divide\") print(\"5.Exponent\") print(\"6.Remainder\") choice = input(\"Enter choice(1/2/3/4/5/6):\") num1 = int(input(\"Enter first number: \")) num2 = int(input(\"Enter second number: \")) if choice == \'1\': print(num1,\"+\",num2,\"=\", add(num1,num2)) elif choice == \'2\': print(num1,\"-\",num2,\"=\", subtract(num1,num2)) elif choice == \'3\': print(num1,\"*\",num2,\"=\", multiply(num1,num2)) elif choice == \'4\': print(num1,\"/\",num2,\"=\", divide(num1,num2)) elif choice == \'5\': print(num1,\"**\",num2,\"=\", exp(num1,num2)) elif choice == \'6\': print(num1,\"%\",num2,\"=\", mod(num1,num2)) else: print(\"Invalid input\") .

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write a topic about RFID in SCM and what you have learned about adoption of RFID in SCM Solution RFID(Radio frequency identification) in SRM is a technology that incorporates the use of electromagnetic or electrostatic coupling in the radio frequency portion of the electromagnetic spectrum to uniquely identify an object, animal, or person. RFDI is coming into increasing use in industry as an alternative to the barcode. The advantage of RFDI is that it does not require direct contact or line of sight scanning. It consists of three components - 1) Antenna, 2. Transreciever, 3. Transponder The term supply chain covers all possible processes involved in the flow of goods from manufacturing to customers including manufacturing, distribution and transportaion. RFDI technology is risen to become a revolutionary elament in supply chain management. It is not replacement for barcodes. RFDI is an innovative technology that will make the supply chain faster and efficient with less human error. There is a common trend when integrating the supply chain for many companies and provides a good model for better understanding of what the adoption of RFDI technology can do for supply chain network. there are many factors that maybe considered for adoption in RFDI in SCM like information technology - complexity, organization compatibility. Industrial organization - firm performance is enabled or constrained by industry structure. Resource based view- presence of resource that meet certain conditions, such as value, rarity, imperfect, imitability and lack of substituability. .

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Write up a simple c-program to find the median of any array? Solution #include <stdio.h> #include <conio.h> void sort (int[],int); float median (int A[], int nElem); int main() {int A[14]={1,2,3,4,4,2,2,1,2,4,1,4,4,4},nElem=14,i,m; float fm; printf(\"The array:\ \"); for(i=0;i<nElem;i++) printf(\"%d \",A[i]); printf(\"\ \ \"); sort(A,nElem); printf(\"The array sorted is:\ \"); for(i=0;i<nElem;i++) printf(\"%d \",A[i]); printf(\"\ \ \"); fm=median(A,nElem); printf(\"The median is %.2f\ \",fm); getch(); return 0; } void sort (int A[], int nElem) {int i,j,t; for(i=0;i<nElem-1;i++) for(j=i+1;j<nElem;j++) if(A[i]>A[j]) {t=A[i]; A[i]=A[j]; A[j]=t; } } float median (int A[], int nElem) {float m; if(nElem%2==0) return((float)(A[nElem/2-1]+A[nElem/2])/2.); else return A[nElem/2]; } .

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Write an assembly language program in the Pep/8 simulator that corresponds to the following C++ program: #include <iostream> using namespace std; int num1; int num2; int main() { cin >> num1 >> num2; cout << num2 << endl << num1 << endl; return 0; } Solution c++ Program: #include<iostream> using namespace std; int num1; int num2; int main() { cin>>num1>>num2; cout<<num2<<endl<<num1<<endl; return 0; } Assembly Language: BR main Sum1: .WORD 0x0000 Sum2: .WORD 0x0000 Num1: .BLOCK 2 Num2: .BLOCK 2 Main: LDA sum1,sum2,d ; â€˜dâ€™ = direct DECI num1,d ADDA num1,d STA sum1,d DEC2 num2,d STA sum2,d DECO sum1,d DECO sum2,d STOP .END .

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Write the for structure in JAVA coding to read and display all elements in the third row of array x. Solution import java.util.Scanner; public class Program1 { public static void main(String[] args) { Scanner sc= new Scanner(System.in); System.out.print(\"Enter number of elements: \"); int n= sc.nextInt(); //creating an array of element n int []x = new int[n]; //now reading array elements System.out.println(\"Enter \"+n+\" elements: \"); for(int i=0; i<n; i++) x[i] = sc.nextInt(); // now displaying using enhanced for loop System.out.println(); System.out.println(\"Array elemets are: \"); for(int e: x) System.out.print(e+\" \"); System.out.println(); } } /* Sample Output: Enter number of elements: 6 Enter 6 elements: 4 8 9 23 12 15 Array elemets are: 4 8 9 23 12 15 */ .

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Write the formal description of the following state machine (M) What is the language recognized by M;L(M)? Solution the language that is accepted is: 0 * 10 * 10 * explanation: start node contains the self loop with value \'0\' i.e it may iterate as many times as it needed zero or more number of times * represents zero or more number of times \' s 0 \' \': it must pass through \'1\' no other choice it reaches s 1 when it is on s 1 it has self loop with \'0\' so it will iterate zero or more number of times i.e 0 * \' s 1 \' \': it must pass through \'1\' no other choice it reaches s 2 when it is on s2 it has self loop with \'0\' so it will iterate zero or more number of times i.e 0 * s 2 is the end state so we cannot go from here an further .

Write the formal description of the following state machine (M) What.docx

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Write the C++ code for a function getInput which will read in an unknown of values from a file and store them in a one-dimensional array of integers. The function stores, in a reference parameter, the actual number of values it reads. The input file is declared and opened in main(), so the input filestream variable is needed as a function parameter. Write a C ++ function to find the minimum in a designated row of a 2-dimensional array which has 10 columns. The function parameters are the array, the row index of the row in which the minimum is to be found, and a reference parameter for the minimum value. Solution 4. #include<iostream.h> #include<conio.h> void min_row(int a[4][10],int row) { int m=a[row][0]; for(int i=1;i<10;i++) { if(a[row][i]<m) m=a[row][i]; } cout<<\"minimum number in row \"<<row<<\" is \"<<m; } void main() { int arr[4][10],i,j,r; clrscr(); cout<<\"Enter Element to the matrix\"; for(i=0;i<4;i++) { cout<<\" Row \"<<i<<\" :\"; for(j=0;j<10;j++) { cout<<\"Column \"<<j<<\":\"; cin>>arr[i][j]; } } cout<<\"\ Enter the row number\"; cin>>r; min_row(arr,r); getch(); } .

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Write the balanced reaction where thiosulfate and protons are the only.docx

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Write functions odd and even, which takes a list of symbols L, and produces a sublist of L containing symbols located at odd and even positions repectively. E.g., (odd \'(a b c d)) = (a c) (even \'(a b c d)) = (b d) (odd \'(a)) = (a) (even \'(a)) = () Solution //just assumed that the size of array is 10 you can chnage it accordingly void oddAndEven(char arr[10]) { char sublistodd[5],sublisteven[5]; int j,k; for(int i=0; i <10 ; i++) { if(i/2 == 0 && i<=4) { sublisteven[i] = arr[i]; j = i; j=j+1; } else if(i/2 != 0 && i<=4) { sublistodd[i] = arr[i]; k = i; k++; } else if(i/2 == 0 && i>4 && j<5) { sublisteven[j] = arr[i]; j++; } else if(i/2 !=0 && i>4 && k<5) { sublistodd[k] = arr[i]; k++; } } } .

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Write about the two sources of energy Write about the two sources of energy Solution Two sources of energy can be solar energy and wind energy Solar Energy Sun is the primary source of all energy. Light given by Sun, sunlight, is a clean and renewable source of energy. It is a sustainable resource which means it doesn\'t run out, but it can be maintained because the sun is always giving off energy. Coal or gas are not categorized as sustainable or renewable because once they are used, there is none left. More and more people have been using clean, renewable energy such as solar, wind, geothermal steam and others. It is called \'Green Power\' . Some Advantages of Solar Energy Some Disadvantages of Solar Energy Wind Energy Another source of renewable energy is wind. Wind can be described as the natural movement of air across the land or sea. To produce electricity, wind is used to turn the blades of a wind mill turns the shaft to which they are attached. This movement of the windmill shaft through a pump or generator produces electricity. Advantages Disadvantages .

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