ACC 315 6-2 Problem Set Question 5 of 7
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ACC 315 6-2 Problem Set Question 5 of 7 (a) Your Answer Correct Answer v Your answer is correct. Transform the general journal above into a single, two-dimensional, flat-file table with no empty cells and only one value in each cell. Abbrevia explanations so that they fit neatly in each cell in the explanation column. Call this flat file of similar records Table 1. (Credit account titles are au indented when the amount is entered. Do not indent manually. If no entry is required, select "No Entry" for the account titles and enter O for the amount. entries before credit entries.) Table 1: General Journal Entries JE_No JE_Date Account_Name Account_No Debit 1. | 7/1/20XX V‘ | Cash | | 1001 Vl ‘ 5000| 1. | 7/1/20XX V' | Accounts Receivable | l 1004 V| | 70000| 1. I 7/1/20XX v] I Sales l [ 4001 vl [ ol 2. | 7/1/20XX v' LCostofGoodsSold | I 5001 v| I 4oooo| 2. | 7/1/20XX v' | Inventory | I 1030 v| I ol 3. | 7/2/20XX V' | Accounts Payable | I 2001 V| | 8000| 3. r7/2/20XX :| rCash ] r1001 :I [ O—I
(a) Your Answer | Correct Answer v Your answer is correct. Transform the general journal above into a single, two-dimensional, flat-file table with no empty cells and only one value in each cell. Abbreviate the explanations so that they fit neatly in each cell in the explanation column. Call this flat file of similar records Table 1. (Credit account titles are automatically indented when the amount is entered. Do not indent manually. If no entry is required, select "No Entry” for the account titles and enter O for the amounts. List all debit entries before credit entries.) Table 1: General Journal Entries Account_Name Account_No Debit Credit Explanation (= | [y | ] | o] [[cmme 9 e | [ | o] | o] [ l Sales ‘ | 4001 vl | o‘ ‘ 75000‘ l Credit sale v‘ l Cost of Goods Sold ‘ | 5001 "' | 40000 l ‘ o ‘ ‘ COGs V‘ = | [ | o | o] [os ‘ Accounts Payable ‘ | 2001 VI | 8000 l ‘ o ‘ ‘ Paid invoice V‘ ‘ Cash ‘ | 1001 vI | o] ‘ eooo‘ Paid invoice V‘ (c1) Your Answer Correct Answer ¥ Your answeris correct. Break down the data into logical, smaller, more manageable units to simplify the organization of data for relational database purposes. Do this by dividing your Table 1 into three tables (Tables A, B, and C), assuming the following additional business rules: « Noattribute in a column must be dependent on only a portion of the unique identifier for a row. Each attribute must be dependent on the entire identifier in your new tables. e Thevalueof anattribute cannot be deduced from the value of another attribute. Hint: Collapse two columns into one to resolve this issue. Prepare Table A. Table A: Transaction Listing JE_Number JE_Date Explanation 1 meoxx j Lmdmsa\cwnhdownpaymem j 2 | 7/1/20XX v‘ l Cost of goods sold 3, ] 7/2/20XX V‘ ‘ Payment of invoice
(c2) Your Answer | Correct Answer v Your answer is correct. Break down the data into logical, smaller, more manageable nits to simplify the organization of data for relational database purposes. Do this by dividing your Table 1into three tables (Tables A, B, and C), assuming the following additional business rules: © Noattribute in a column must be dependent on only a portion of the unique identifier for a row. Each attribute must be dependent on the entire identifier in your new tables. . The value of an attribute cannot be deduced from the value of another attribute. Hint: Collapse two columns into one to resolve this issue. Prepare Table B. Table B: Transaction Details JE_Number JE_Date Account_No Amount 1 [7amo <] [[1001 ¥ [ 5000 | 1 l 7/1/20XX V‘ l 1004 v‘ [ 70000| 1 l 7/1/20XX v‘ l 4001 v‘ [ 75000 n 2. l 7/1/20XX v‘ l 5001 v‘ [ 40000| 2. l 7/1/20XX v‘ l 1030 v‘ [ 40000 n 3. ‘ 7/2/20%X v‘ l 2001 v‘ [ goool 3 ‘ 7/2/20XK v‘ l 1001 V‘ [ 8000 n
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(c3) Your Answer | Correct Answer v Your answer is correct. Break down the data into logical, smaller, more manageable units to simplify the organization of data for relational database purposes. Do this by dividing your Table 1into three tables (Tables A, B, and C), assuming the following additional business rules: . No attribute in a column must be dependent on only a portion of the unique identifier for a row. Each attribute must be dependent on the entire identifier in your new tables. The value of an attribute cannot be deduced from the value of another attribute. Hint: Collapse two columns into one to resolve this issue. Prepare Table C. Table C: Chart of Accounts Account No Account Name l 1001 V‘ l Cash i [ e] [ecomsree | l 1030 V‘ l Inventory I ‘ 2001 V‘ l Accounts Payable I ‘ 4001 V‘ ‘ Sales I ‘ 5001 V‘ ‘ Cost of Goods Sold I
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