If the sample means are different from the population mean, what do .pdf

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mayorothenguyenhob69

I have a generator and I need a wheel that is going to start the generator. How I know the diameter of the wheel that I need to start the engine. The require rpm for start of the generator is 3480-3780. Solution Basically there are many ways to get the dia of wheel but one method is by using the flywheel of corresponding power the flwheel size define the wheel size of the used in generator by using the fluctuation in speed as given.

If the sample means are different from the population mean, what do we call this discrepancy?
Solution
Sample mean for large number of samples tend to follow a normal distribution with mean =
sample mean, and deviation from population mean measured as square
The standard error is the square root of sum of deviations of sample means from mean/ square
root of each sample size
Hence the discrepancy is called std error.

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Compare different sexual and asexual life cycles noting their adaptive advantages. Solution Asexual reproduction sexual reproduction Number of organisms involved One parent needed Two parents are required to mate Cell division Cells divide by Fission, budding , or regeneration Cells divide by Meiosis Types Budding, vegetative reproduction, fragmentation, spore formation Syngamy and conjugation Advantages Time Efficient; no need to search for mate, requires less energy Variation, Unique., organism is more protected Disadvantages No variation - if the parent has a genetic disease, offspring does too. Requires two organisms, requires more energy Evolution There is very little chance of variation with asexual reproduction. Mutations in DNA can still occur but not nearly as frequently as in sexual reproduction. Sexual reproduction leads to genetic variation in new generations of offspring. This is fundamental to evolution. Involvement of sex cells No formation or fusion of gametes (sex cell) Formation and fusion of gametes (sex cell) occurs Found in Lower organisms Higher invertebrates and all vertebrates Unit of reproduction May be whole parent body or a bud or a fragment or a single somatic cell Gamete Time taken Asexual reproduction is completed in a very short period of time. Sexual reproduction can take several months to complete. Number of offspring Two or more One or more.

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Background: In many applications, the composition of a collection of data items changes over time. Not only are new data items added and existing ones removed, but data items may be duplicated. A list data structure is a member of the general category of abstract data types called containers, whose purpose is to hold other objects. In C++, lists are provided in the Standard Template Library. However, for this assignment you will design and write your own linked list implementation to support the ADT operations specified below. Objective: Design and implement the specifications for a List Abstract Data Type where the items in the list are unsorted. Requirements: Define a list and develop a set of operations for creating and manipulating a list that satisfies the list ADT specification. List ADT Specification Structure: The list elements are of ItemType. The list has a property called the current position which designates the position of the last element accessed by GetNextItem during an iteration through the list. Only ResetList and GetNextItem alter the current position. Definitions (provided by the user): MAX_ITEMS: A constant specifying the maximum capacity of items allowed on the list Item Type: Class encapsulating the type of items in the list RelationType: An enumeration type that consists of LESS, GREATER, EQUAL Member function of ItemType that must be included: RelationType ComparedTo(ItemType Item) Function: Determines the ordering of two ItemType objects based on their keys Precondition: Self and item have their key members initialized Postcondition: Function value = LESS if the key of self is less than the key of item = GREATER if the key of self is greater than the key of item = EQUAL if the keys are equal Operations (provided by Unsorted List ADT) Make Empty Function: Initializes list to empty state Preconditions: None Postcondition: List is empty Boolean IsFull Function: Determines whether list is full Preconditions: List has been initialized Postcondition: Function value = (list is full) int GetLength Function: Determines the number of elements in list Preconditions: List has been initialized Postcondition: Function value = number of elements in list ItemType GetItem(Item Typeitem, Boolean& found) Function: Get list element whose key matches item’s key (if present) Preconditions: List has been initialized Key member of item is initialized Postcondition: If there is an element someItem whose keymatches item’s key, then found = true and copy of someItemis returned; otherwise found = false and item is returned List is unchanged PutItem(ItemType item) Function: Puts item into list Preconditions: List has been initialized List is not full Item is not in list Postcondition: Item is in the list DeleteItem(ItemType item) Function: Deletes the element whose key matches item’s key Preconditions: List has been initialized Postcondition: One and only one element in list has a key matching item’s key ResetList Function: Initializes current p.

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mayorothenguyenhob69

11.25 In 2010, there were 117,538,000 households in the United States. There were 78,833,000 family households made up of married couples, single-male households, and single-female households. To determine how many of each type a survey was undertaken. The results were stored using the codes 1=married couples, 2=single male, and 3=single female. Estimate with 95% confidence the total number of American households with married couples. (Adapted from Statistical Abstract of the United States, 2012, Table 59) Solution.

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mayorothenguyenhob69

What is the first enzymatic activity in DNA synthesis? Endonuclease Primase Exonuclease DNA polymerase LigaseA. EndonucleaseB. PrimaseC. ExonucleaseD. DNA polymeraseE. Ligase Solution Ans. Correct option. B. Primase. The first enzymatic activity in DNA synthesis (among the available options) is Primase. DNA primase (or, primase) binds to ssDNA strands during initiation period. It synthesizes small RNA primer to both the strands. DNA polymerase holoenzyme binds during elongation of DNA synthesis after primase has synthesized RNA primers. It extends the RNA primers with DNA nucleotide polymerization on it in 5’-3’ direction. The exonuclease activity of DNA polymerase comes into action during repair or nick translation during elongation. Other than that, no other endonuclease or exonuclease participate DNA synthesis. Ligase (DNA ligase) joins DNA fragments during nick translation much later than DNA pol begins to synthesize DNA..

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What is the probability that a five card poker hand contains the jack of diamonds and the two of spades? Solution Jack of diamonds and the two of spades: No of ways to get 1 Jack of diamonds: = 1 No of ways to get 2 spades: 13C2 = 78 Total No of ways to contains the jack of diamonds and the two of spades = 1*78 = 78 and No of ways to five card poker hand = 52C5 = 2,598,960 Then Probability = 78/2,598,960 = 0.00003001.

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mayorothenguyenhob69

Use the following information to answer questions 7-11 The following information applies to the questions displayed below. The following information is available for Lock-Tite Company, which produces special-order security products and uses a job order costing system. April 30 May 31 Inventories Raw materials Work in process Finlshed goods $39,000 $30,000 9,100 20,500 69,000 34,500 Activities and information for May Raw materials purchases (paid with cash) Factory payroll (paid with cash) Factory overhead 186,000 300,000 Indirect materials Indirect labor Other overhead costs 7,000 71,000 111,000 1,100,000 Sales (received in cash) Predetermined overhead rate based on direct labor cost 55% Solution Answer - Exercise 15-7 T-accounts Factory Overheads Cost of material used Raw Materials Raw Material 7000 125950 Work in process Raw Materials April-30 $39,000 RM-April 30 39000 7000 Factory Overhead Factory Payroll 71000 63050 Cost of goods sold Add:Raw Material Purchases $186,000 RM Purchases 186000 188000 Work in process Other Overheads 111000 Less : Raw Materials May-31 $30,000 Cost of material used $195,000 less : Indirect Material used $7,000 RM May 31 30000 Direct Material Used $188,000 Work in process Cost of direct labour used = $3,00,000 WIP- April 30 9100 531550 Finished Goods Raw Material 188000 Cost of goods Manufactured Factory Payroll 229000 Cost of Materials used $188,000 Factory Overhead 125950 Cost of direct labour used $229,000 WIP-May 31 20500 Factory Overhead applied [55% of $2,29,000] $125,950 Cost of Manufacturing for the period $542,950 Factory Payroll Add: Work in process , April-30 $9,100 Cash 300000 71000 Factory Overheads Less : Work in process,May-31 $20,500 229000 Work in process Cost of goods Manufactured $531,550 Cost of goods sold Cost of goods Manufactured $531,550 Add : Finished Goods , April-30 $69,000 Finished Goods Less : Finished Goods,May-31 $34,500 FG-April 30 69000 Cost of goods sold $566,050 Work in process 531550 566050 Cost of goods sold Gross Profit Sales $1,100,000 FG May 31 34500 Less : Cost of goods sold (unadjusted) $566,050 Gross Profit $533,950 Overapplied or underapplied overheads = Actual overheads - applied overheads Actual Overheads = Indirect Material + Indirect Labour + other overhead costs = $7000 + $71000 + $111000 = $1,89,000 Underapplied overheads = $189000 -$125950 = $63,050 Journal Entry to allocate underapplied overheads Account Titles Debit Credit Cost of goods sold $63,050 Factory Overheads $63,050 Income Statement (Partial) Sales $1,100,000 Cost of goods sold $566,050 Gross Profit $533,950.

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Use Simplex method and solve the following problem: Maximize p = x+y subject to x+y <= 2, 3x+y >= 4 Which of teh following is the last matrix table? A. x y s1 s2 | b 0 1 3/2 1/2 1 1 0 1/2 -1/2 1 0 0 1 0 2 B. x y s1 s2 | b 0 1 3/2 1/2 1 1 0 -1/2 1/2 1 0 0 1 0 2 C. x y s1 s2 | b 0 1 3/2 1/2 1 1 0 -1/2 -1/2 1 0 0 1 0 2 D. x y s1 s2 | b 0 1 3/2 -1/2 1 1 0 -1/2 -1/2 1 0 0 1 0 2 x y s1 s2 | b 0 1 3/2 1/2 1 1 0 1/2 -1/2 1 0 0 1 0 2 B. x y s1 s2 | b 0 1 3/2 1/2 1 1 0 -1/2 1/2 1 0 0 1 0 2 C. x y s1 s2 | b 0 1 3/2 1/2 1 1 0 -1/2 -1/2 1 0 0 1 0 2 D. x y s1 s2 | b 0 1 3/2 -1/2 1 1 0 -1/2 -1/2 1 0 0 1 0 2 Solution The step by step solution of the problem is given below - Tableau #1 x y s1 s2 p 1 1 1 0 0 2 3 1 0 -1 0 4 -1 -1 0 0 1 0 Tableau #2 x y s1 s2 p 0 2/3 1 1/3 0 2/3 1 1/3 0 -1/3 0 4/3 0 -2/3 0 -1/3 1 4/3 Tableau #3 x y s1 s2 p 0 1 3/2 1/2 0 1 1 0 -1/2 -1/2 0 1 0 0 1 0 1 2 So, the correct answer is option (c)..

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There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user\'s magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, what is the critical value for the hypothesis test? z = z = Solution It is a two-tailed test. Given a=0.05, the critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table) Answer: z =.

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think like an You have three resistors of resistance 30 ohm [Ohm]. There are__configurations possible, with__unique combinations, resulting in___different resistance values. A \"configuration\" is a way of combining the resistors. For example, two resistors in parallel is one configuration; two resistors in series is a second configuration. A \"combination\" is the specific way of combining given resistors to form an effective resistance. For example, combining Resistor #1 and #2 in parallel is one combination; combining Resistor #1 and #3 in parallel is a second combination. These combinations may or may not result in a unique effective resistance. What is the greatest resistance that can be made from a combination of resistors, and what is the effective resistance of this combination? What is the least resistance that can be made from a combination of resistors, and what is the effective resistance of this combination? Solution b) GREATEST RESISTANCE CAN BE ACHIEVED THROUGH SERIES CONNECTION OF THREE RESISTORS. EFFECTIVE RESISTANCE=30+30+30=90 OHMS C) LEAST RESISTANCE CAN BE ACHIEVED THROUGH PARALLEL CONNECTION OF THREE RESISTORS. EFFECTIVE RESISTANCE=(30*30)/(30+30)=15 OHMS (15*30)/(15+30)=450/45=10 OHMS.

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mayorothenguyenhob69

30-32 Nuclear pores permit the passage of chromosomes outward. glucose molecules outward. assembled DNA molecules outward. RNA outward. sodium ions inward, potassium ions outward. The cell walls of fungi are made from peptidoglycan. pseudomurein. cellulose. chitin. Starch. Flattened sacks of membranes involved in the packaging and distribution of molecules synthesized in the cell are known as Golgi bodies. microbodies. pinocytic vesicles. vacuoles. chromosomes. Solution 30). RNA outwarrd This transport includes RNA and ribosomal proteins moving from nucleus to the cytoplasm. 31). Cellulose and chitin 32). Golgi bodies.

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Q3 GIVEN THE TREND EQUATION NO OF STUDENTS-3000+200 YEAR BASE: 2003-0 1.INTERPRET EQUATION RESULTS 2.FORECAST FOR 2020 3.WHEN WILL THE NO OF STUDENTS REACH 6000 Solution x no of years from 2003 y no of students y = 3000+200x 1) The regression equation has a positive slope of 200 y intercept is 3000 i.e. no of students in the year 2000 Thus for every year 200 students increase 2) For 2020, x = 17, y = 3000+200(17) = 6400 3) When y =6000, 6000 = 3000+200x or x = 15 Hence in year 2018 students reach 6000 B Regression C D. All the above.

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B-tree insertion (Draw node) Show the results of inserting the keys 23, 98, 55, 37, 30, 47, 35, 15, 10, 75, 3, 53, 1, 33, 5, 28, 25, 90, 95, 56, 8, and 9 in order into an empty B-tree of order 5. That is, each internal node in this B-tree other than the root must have at least 3 and at most 5 children (or at least 2 and at most 4 keys). Solution /* * C++ Program to Implement B-Tree */ #include #include #include using namespace std; struct BTreeNode { int *data; BTreeNode **child_ptr; bool leaf; int n; }*root = NULL, *np = NULL, *x = NULL; BTreeNode * init() { int i; np = new BTreeNode; np->data = new int[5]; np->child_ptr = new BTreeNode *[6]; np->leaf = true; np->n = 0; for (i = 0; i < 6; i++) { np->child_ptr[i] = NULL; } return np; } void traverse(BTreeNode *p) { cout<n; i++) { if (p->leaf == false) { traverse(p->child_ptr[i]); } cout << \" \" << p->data[i]; } if (p->leaf == false) { traverse(p->child_ptr[i]); } cout< p[j]) { temp = p[i]; p[i] = p[j]; p[j] = temp; } } } } int split_child(BTreeNode *x, int i) { int j, mid; BTreeNode *np1, *np3, *y; np3 = init(); np3->leaf = true; if (i == -1) { mid = x->data[2]; x->data[2] = 0; x->n--; np1 = init(); np1->leaf = false; x->leaf = true; for (j = 3; j < 5; j++) { np3->data[j - 3] = x->data[j]; np3->child_ptr[j - 3] = x->child_ptr[j]; np3->n++; x->data[j] = 0; x->n--; } for (j = 0; j < 6; j++) { x->child_ptr[j] = NULL; } np1->data[0] = mid; np1->child_ptr[np1->n] = x; np1->child_ptr[np1->n + 1] = np3; np1->n++; root = np1; } else { y = x->child_ptr[i]; mid = y->data[2]; y->data[2] = 0; y->n--; for (j = 3; j < 5; j++) { np3->data[j - 3] = y->data[j]; np3->n++; y->data[j] = 0; y->n--; } x->child_ptr[i + 1] = y; x->child_ptr[i + 1] = np3; } return mid; } void insert(int a) { int i, temp; x = root; if (x == NULL) { root = init(); x = root; } else { if (x->leaf == true && x->n == 5) { temp = split_child(x, -1); x = root; for (i = 0; i < (x->n); i++) { if ((a > x->data[i]) && (a < x->data[i + 1])) { i++; break; } else if (a < x->data[0]) { break; } else { continue; } } x = x->child_ptr[i]; } else { while (x->leaf == false) { for (i = 0; i < (x->n); i++) { if ((a > x->data[i]) && (a < x->data[i + 1])) { i++; break; } else if (a < x->data[0]) { break; } else { continue; } } if ((x->child_ptr[i])->n == 5) { temp = split_child(x, i); x->data[x->n] = temp; x->n++; continue; } else { x = x->child_ptr[i]; } } } } x->data[x->n] = a; sort(x->data, x->n); x->n++; } int main() { int i, n, t; cout<<\"enter the no of elements to be inserted\ \"; cin>>n; for(i = 0; i < n; i++) { cout<<\"enter the element\ \"; cin>>t; insert(t); } cout<<\"traversal of constructed tree\ \"; traverse(root); getch(); }.

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mayorothenguyenhob69

OPERATING SYSTEMS: DISK SCHEDULING PLEASE SHOW YOUR WORK! I\'M TRYING TO UNDERSTAND HOW TO ACTUALLY DO THE PROBLEM. 2. (3 points) Suppose that a disk drive has 3000 cylinders, numbered 0 to 2999. The drive is currently serving a request at cylinder 140, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is 66, 147, 913, 289, 202, 1750, 15 Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms? a. FCFS b. SSTF c. SCAN d. LOOK e. C-SCAN 3. Requests are not usually uniformly distributed. For example, a cylinder containing the file system FAT or inodes can be expected to be accessed more frequently than a cylinder that only contains files. Suppose you know that 50 percent of the requests are for a small, fixed number of cylinders. Propose a disk-scheduling algorithm (one that we have discussed) that gives even better performance by taking advantage of this “hot spot” on the disk. Explain your answer and discuss any problems and how you might mitigate them. Solution The drive is currently serving a request at cylinder 140 request queue -- 66,147,913,289,202,1750,15 a) FCFS 66<--- 140 66---->147 147 ----> 913 289 <----- 913 202<---- 289 202 ------>1750 15 <------ 1750 total time = |66-140|+|66-147|+|147-913|+|289-913|+||202-289|+|202-1750|+|15-1750| =74+81+766+624+87+1548+1735 =4915 b) SSTF 140 ----> 147 147 ------>202 202----->289 66<----289 15<----66 15----->913 913----->1750 total time=7+55+87+223+51+898+837 =2158 c) SCAN 140--->147---->202---->289----->913----->1750----->2999--- 0<-----15<------66<----------------------------------------------- total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-66|+|66- 15|+|15-0| =7+55+87+624+837+1249+2933+51+15 =5858 d)LOOK 140--->147---->202---->289----->913----->1750--- 15<------66<----------------------------------------------- total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-66|+|66-15| =7+55+87+624+837+1684+51 =3345 e)C-SCAN 140--->147---->202---->289----->913----->1750----->2999--- 66<-----15<------0<----------------------------------------------- total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-0|+|15- 0|+|66-15| =7+55+87+624+837+1249+2999+15+51 =5924 SSTF can be the best disk scheduling algorithm when data is close to each other. We can have hot spot in the middle of the disk . Aging can be done on older request by increasing their priority.

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Joe\'s Coffee Shoppe orders fresh doughnuts each morning. Joe pays $3.20 per dozen doughnuts and sell doughnuts for $0.4 each. The leftover doughnuts are salvaged at half the selling price. Suppose that the demand for doughnuts is uniformly distributed from 16 dozens to 30 dozens. How many doughnuts (in dozens) should Joe stock each day? A detailed analysis of past data shows that the number of doughnuts sold per day is better described by a normal distribution, with a mean 23 dozens and standard deviation 10 dozens. Now, how many doughnuts should Joe stock each day? The expected demand was the same in parts (a) and (b), but the optimal order quantities should have been different. What accounted for this different. What accounted for this difference? Solution.

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In a survey conducted by a business management company, 51% of the U.S. workers siad they have negotiated a pay raise at least once inthier lives. The sample size used in the study was not disclosed. A) Suppose that the survey had a sample size of n=500. Construct a condifidence interval estimate for the proportion of all United States workers who have negotiated a pay raise. Solution Answer to the question) P = 0.51 n = 500 Standard error SE = srt(0.51 *0.49/500) SE = 0.0224 . Let the confidence level = 95% , so the Z critical value = 1.96 . The formula of cofndience interval is: p - z* SE , p + z * SE . On plugging the values we get 0.51 - 1.96 *0.0224 , 0.51 +1.96 *0.0224 0.04661 , 0.5539.

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