explain please why you read an articles about
relational database management system.
why it is important to understand the characteristics of relational
database management system?
In: Computer Science
In: Computer Science
Can someone please explain graph database countermeasures? I tried a lot to look online, but did not find anything regarding countermeasures for graph database.
In: Computer Science
Discuss ONE (1) of the following ethical dilemmas using the ethical decision-making framework (model) given below or another model that you know I’m working on a project that will make a lot of people in our business redundant. Soon we will prepare the list of people affected and I know my friends will be included. I feel terrible and am not sure I can keep working on it. What can I do?
In: Accounting
Andies Co. held 80% of the common stock of Breaver Inc. and 40% of this subsidiary's convertible bonds. the following consolidated financial statements were for 2010 and 2011.
| 2010 | 2011 | |
| Revenues | $ 1,064,000 | $ 1,232,000 |
| Cost of goods sold | (714,000) | (756,000) |
| Depreciation and amortization | (126,000) | (140,000) |
| Gain on sale of building | - | 28,000 |
| Interest expense | (42,000) | (42,000) |
| Non-controlling interest | 12,600 | 15,400 |
| Net income to controlling interest | 169,400 | 306,600 |
| Retained earnings, January 1 | $ 420,000 | $ 519,400 |
| Net income (from above) | 169,400 | 306,600 |
| Dividends paid | (70,000) | (140,000) |
| Retained earnings, December 31 | 519,400 | 686,000 |
| Cash | $ 112,000 | $ 196,000 |
| Accounts receivable | 210,000 | 196,000 |
| Inventory | 280,000 | 476,000 |
| Buildings and equipment (net) | 896,000 | 966,000 |
| Database | 210,000 | 203,000 |
| Total assets | 1,708,000 | 2,037,000 |
| Accounts payable | $ 196,000 | $ 140,000 |
| Bonds payable | 560,000 | 720,000 |
| Non-controlling interest | 44,800 | 57,400 |
| Common stock | 140,000 | 168,000 |
| Additional paid-in capital | 247,800 | 265,600 |
| Retained earnings, December 31 | 519,400 | 686,000 |
| Total liabilities and stockholders' equity | 1,708,000 | 2,037,000 |
Additional Information:
1. Bonds were issued during 2011 by the parent for cash.
2. Amortization of a database acquired in the original combination amounted to $7,000 per year.
3. A building with a cost of $84,000 but a $42,000 book value was sold by the parent for cash on May 11, 2011.
4. Equipment was purchased by the subsidiary on July 23, 2011, using cash.
5. Late in November 2011, the parent issued common stock for cash.
6. During 2011, the subsidiary paid dividends of $14,000.
Required:
Prepare a consolidated statement of cash flows (indirect method) for this business combination for the year ending December 31, 2011.
In: Accounting
Andies Co. held 80% of the common stock of Breaver Inc. and 40% of this subsidiary's convertible bonds. The following consolidated financial statements were for 2010 and 2011.
| 2010 | 2011 | |
| Revenues | $1,064,000 | $1,232,000 |
| Cost of goods sold | (714,000) | (756,000) |
| Depreciation and amortization | (126,000) | (140,000) |
| Gain on sale of building | 0 | 28,000 |
| Interest expense | (42,000) | (42,000) |
| Non-controlling interest | 12,600 | 15,400 |
| Net income to controlling interest | $169,400 | $306,600 |
| Retained earnings, Jan 1 | $420,000 | $519,400 |
| Net income (from above) | 169,400 | 306,600 |
| Dividends paid | (70,000) | (140,000) |
| Retained earnings, Dec 31 | $519,400 | $686,000 |
| Cash | $112,000 | $196,000 |
| Accounts receivable | 210,000 | 196,000 |
| Inventory | 280,000 | 476,000 |
| Buildings and equipment (net) | 896,000 | 966,000 |
| Database | 210,000 | 203,000 |
| Total assets | $1,708,000 | $2,037,000 |
| Accounts payable | $196,000 | $140,000 |
| Bonds payable | 560,000 | 720,000 |
| Non-controlling interest | 44,800 | 57,400 |
| Common stock | 140,000 | 168,000 |
| Additional paid-in capital | 247,800 | 265,600 |
| Retained earnings, Dec 31 | 519,400 | 686,000 |
| Total liabilities and stockholders' equity | $1,708,000 | $2,037,000 |
Additional information:
1. Bonds were issued during 2011 by the parent for cash.
2. Amortization of a database acquired in the original combination amounted to $7,000 per year.
3. A building with a cost of $84,000 but a $42,000 book value was sold by the parent for cash on May 11, 2011.
4. Equipment was purchased by the subsidiary on July 23, 2011, using cash.
5. Late in November 2011, the parent issued common stock for cash.
6. In 2011, the subsidiary paid dividends of $14,000.
Required:
Prepare a consolidated statement of cash flows (indirect method) for this business combination for the year ending December 31.
In: Accounting
1. Write the SQL code required to list the employee number, first and last name, middle initial, and the hire date. Sort your selection by the hire date, to display the newly hired employees first.
2. Modify the previous query and list the employee first, last name and middle initial as one column with the title EMP_NAME. Make sure the names are listed in the following format: First name, space, initial, space, last name (e.g. John T Doe). Hint: use + to concatenate the fields and use ' ' to add a space to your text. E. g. EMP_FNAME + ' ' + EMP_LNAME.
3. Write the SQL code that will list only the distinct project numbers in the ASSIGNMENT table, sorted by project number.
4. Using the EMPLOYEE and PROJECT tables, write the SQL code that will join the tables on their common attribute and display the the names and numbers of the projects and the employees who lead these projects.
5. Modify the query in 5 to display all employees and not just the ones who lead projects.
6. Using the EMPLOYEE, JOB, and PROJECT tables in the ConstructCo database, write the SQL code that will join the EMPLOYEE and PROJECT tables using EMP_NUM as the common attribute. Display the attributes shown in the results presented in the attached screen shot, sorted by project value.
7. Using JOB and EMPLOYEE tables, list the jobs and the names of employees who currently have these job categories. List all jobs, even the ones that do not have any matches in the EMPLOYEE table.
8. Write a query to list the names of all employees, the names of the projects to which they are assigned, and the name of employees who lead these projects.
Hint for Question 8: in this question, you will need to join EMPLOYEE to ASSIGNMENT to PROJECT to EMPLOYEE again. Being in the join twice, EMPLOYEE needs aliases:
SELECT ...
FROM EMPLOYEE E1
LEFT JOIN ASSIGNMENT ON ...
LEFT JOIN PROJECT ON ...
LEFT JOIN EMPLOYEE E2 ON ...
When you have aliases, every field from the table should be preceded by the table alias, e.g. SELECT E1.EMP_LNAME, not SELECT EMP_LNAME.
CODE:
/* Database Systems, Coronel/Morris */
/* Type of SQL : SQL Server */
CREATE TABLE ASSIGNMENT (
ASSIGN_NUM int,
ASSIGN_DATE datetime,
PROJ_NUM varchar(3),
EMP_NUM varchar(3),
ASSIGN_JOB varchar(3),
ASSIGN_CHG_HR numeric(8,2),
ASSIGN_HOURS numeric(8,2),
ASSIGN_CHARGE numeric(8,2)
);
INSERT INTO ASSIGNMENT VALUES('1001','3/22/2018','18','103','503','84.5','3.5','295.75');
INSERT INTO ASSIGNMENT VALUES('1002','3/22/2018','22','117','509','34.55','4.2','145.11');
INSERT INTO ASSIGNMENT VALUES('1003','3/22/2018','18','117','509','34.55','2','69.10');
INSERT INTO ASSIGNMENT VALUES('1004','3/22/2018','18','103','503','84.5','5.9','498.55');
INSERT INTO ASSIGNMENT VALUES('1005','3/22/2018','25','108','501','96.75','2.2','212.85');
INSERT INTO ASSIGNMENT VALUES('1006','3/22/2018','22','104','501','96.75','4.2','406.35');
INSERT INTO ASSIGNMENT VALUES('1007','3/22/2018','25','113','508','50.75','3.8','192.85');
INSERT INTO ASSIGNMENT VALUES('1008','3/22/2018','18','103','503','84.5','0.9','76.05');
INSERT INTO ASSIGNMENT VALUES('1009','3/23/2018','15','115','501','96.75','5.6','541.80');
INSERT INTO ASSIGNMENT VALUES('1010','3/23/2018','15','117','509','34.55','2.4','82.92');
INSERT INTO ASSIGNMENT VALUES('1011','3/23/2018','25','105','502','105','4.3','451.5');
INSERT INTO ASSIGNMENT VALUES('1012','3/23/2018','18','108','501','96.75','3.4','328.95');
INSERT INTO ASSIGNMENT VALUES('1013','3/23/2018','25','115','501','96.75','2','193.5');
INSERT INTO ASSIGNMENT VALUES('1014','3/23/2018','22','104','501','96.75','2.8','270.9');
INSERT INTO ASSIGNMENT VALUES('1015','3/23/2018','15','103','503','84.5','6.1','515.45');
INSERT INTO ASSIGNMENT VALUES('1016','3/23/2018','22','105','502','105','4.7','493.5');
INSERT INTO ASSIGNMENT VALUES('1017','3/23/2018','18','117','509','34.55','3.8','131.29');
INSERT INTO ASSIGNMENT VALUES('1018','3/23/2018','25','117','509','34.55','2.2','76.01');
INSERT INTO ASSIGNMENT VALUES('1019','3/24/2018','25','104','501','110.5','4.9','541.45');
INSERT INTO ASSIGNMENT VALUES('1020','3/24/2018','15','101','502','125','3.1','387.5');
INSERT INTO ASSIGNMENT VALUES('1021','3/24/2018','22','108','501','110.5','2.7','298.35');
INSERT INTO ASSIGNMENT VALUES('1022','3/24/2018','22','115','501','110.5','4.9','541.45');
INSERT INTO ASSIGNMENT VALUES('1023','3/24/2018','22','105','502','125','3.5','437.5');
INSERT INTO ASSIGNMENT VALUES('1024','3/24/2018','15','103','503','84.5','3.3','278.85');
INSERT INTO ASSIGNMENT VALUES('1025','3/24/2018','18','117','509','34.55','4.2','145.11');
/* -- */
CREATE TABLE EMPLOYEE (
EMP_NUM varchar(3),
EMP_LNAME varchar(15),
EMP_FNAME varchar(15),
EMP_INITIAL varchar(1),
EMP_HIREDATE datetime,
JOB_CODE varchar(3),
EMP_YEARS int
);
INSERT INTO EMPLOYEE VALUES('101','News','John','G','11/8/2000','502','4');
INSERT INTO EMPLOYEE VALUES('102','Senior','David','H','7/12/1989','501','15');
INSERT INTO EMPLOYEE VALUES('103','Arbough','June','E','12/1/1996','503','8');
INSERT INTO EMPLOYEE VALUES('104','Ramoras','Anne','K','11/15/1987','501','17');
INSERT INTO EMPLOYEE VALUES('105','Johnson','Alice','K','2/1/1993','502','12');
INSERT INTO EMPLOYEE VALUES('106','Smithfield','William','','6/22/2004','500','0');
INSERT INTO EMPLOYEE VALUES('107','Alonzo','Maria','D','10/10/1993','500','11');
INSERT INTO EMPLOYEE VALUES('108','Washington','Ralph','B','8/22/1991','501','13');
INSERT INTO EMPLOYEE VALUES('109','Smith','Larry','W','7/18/1997','501','7');
INSERT INTO EMPLOYEE VALUES('110','Olenko','Gerald','A','12/11/1995','505','9');
INSERT INTO EMPLOYEE VALUES('111','Wabash','Geoff','B','4/4/1991','506','14');
INSERT INTO EMPLOYEE VALUES('112','Smithson','Darlene','M','10/23/1994','507','10');
INSERT INTO EMPLOYEE VALUES('113','Joenbrood','Delbert','K','11/15/1996','508','8');
INSERT INTO EMPLOYEE VALUES('114','Jones','Annelise','','8/20/1993','508','11');
INSERT INTO EMPLOYEE VALUES('115','Bawangi','Travis','B','1/25/1992','501','13');
INSERT INTO EMPLOYEE VALUES('116','Pratt','Gerald','L','3/5/1997','510','8');
INSERT INTO EMPLOYEE VALUES('117','Williamson','Angie','H','6/19/1996','509','8');
INSERT INTO EMPLOYEE VALUES('118','Frommer','James','J','1/4/2005','510','0');
/* -- */
CREATE TABLE JOB (
JOB_CODE varchar(3),
JOB_DESCRIPTION varchar(25),
JOB_CHG_HOUR numeric(8,2),
JOB_LAST_UPDATE datetime
);
INSERT INTO JOB VALUES('500','Programmer', '35.75','11/20/2017');
INSERT INTO JOB VALUES('501','Systems Analyst', '96.75','11/20/2017');
INSERT INTO JOB VALUES('502','Database Designer', '125', '3/24/2018');
INSERT INTO JOB VALUES('503','Electrical Engineer', '84.5', '11/20/2017');
INSERT INTO JOB VALUES('504','Mechanical Engineer', '67.9', '11/20/2017');
INSERT INTO JOB VALUES('505','Civil Engineer', '55.78','11/20/2017');
INSERT INTO JOB VALUES('506','Clerical Support', '26.87','11/20/2017');
INSERT INTO JOB VALUES('507','DSS Analyst', '45.95','11/20/2017');
INSERT INTO JOB VALUES('508','Applications Designer','48.1', '3/24/2018');
INSERT INTO JOB VALUES('509','Bio Technician', '34.55','11/20/2017');
INSERT INTO JOB VALUES('510','General Support', '18.36','11/20/2017');
/* -- */
CREATE TABLE PROJECT (
PROJ_NUM varchar(3),
PROJ_NAME varchar(25),
PROJ_VALUE numeric(10,2),
PROJ_BALANCE numeric(10,2),
EMP_NUM varchar(3)
);
INSERT INTO PROJECT VALUES('15','Evergreen','1453500','1002350','103');
INSERT INTO PROJECT VALUES('18','Amber Wave','3500500','2110346','108');
INSERT INTO PROJECT VALUES('22','Rolling Tide','805000','500345.2','102');
INSERT INTO PROJECT VALUES('25','Starflight','2650500','2309880','107');In: Computer Science
How many structures can you find to use as a starting dataset if you wanted to study how proteins interact with DNA. Assume that you will require only high resolution structures for the study (at least 2.5 Angstroms resolution). For the study, you plan to use structures that were only solved by X-ray crystallography. (Select the closest number if the exact number is not available) *
between 1000-1500
between 2700 - 4000
172
371
between 450 - 950
What is the UniProtKB? *
The UniProt Knowledgebase (UniProtKB) is the central hub for the collection of functional information on proteins, with accurate, consistent and rich annotation.
The UniProt Knowledgebase (UniProtKB) is the universal protein information center.
The UniProtKB is a database of unique protein sequences.
The UniProtKB is a database of conserved protein sequences.
The UniProtKB is a database of protein 3D structures using the Cartesian coordinate format.
Retrieve the database entry for the protein BPSL1549 from the GenBank database. What organism is the protein from? *
Burkholderia pseudomallei
Pseudomonas aeruginosa
Vibrio cholera
Burkholderia mallei
E. coli
Search the GenBank database for the database entry with the following accession number: MN908947. Select from the list below answers that are TRUE statements pertaining to that database entry. *
It is a nucleotide sequence composed of ~29,903 bp
The data is a genome sequence.
The data is the genome sequence of SARS-CoV-2
The data entry is from the order Nidovirales.
The data entry is from the order Coronavirales
The genome sequence was reported by Wuhan et al in the journalNature, 579, published in 2020.
The genome was isolated from patients who had suffered severe breathing difficulties that in some cases were fatal.
The organism caused a bacterial pneumonia that presented as a severe and acute respiratory symptoms.
The sequence was deposited into the GenBank database on 18 May 2020 and is the newest version of the COVID-19 sequence available.
In: Biology
1. As of December 31, 2020, the Russell Corporation has
10,000 shares ($100 PAR) 10% preferred stockissued and outstanding
20,000 shares of common stock ($10 par value) issued and outstanding
Dividends for 2018 and 2019 have NOT been paid – THEREFORE THERE ARE TWO YEARS OFF DIVIDENDS IN ARREARS ON THE PREFERRED STOCK.
At December 31, 2020, Russell Corp. declares total cash dividends of $500,000 to be paid to both the preferred stockholders and the common stockholders. Fill in the table below to indicate the amount of cash that will be distributed to both the preferred stockholders and the common stockholders, on December 31, 2020, under each of the independentsituations
|
Distributed to preferred stockholders |
Distributed to common stockholders |
|
|
Preferred stock is NON cumulative and NON participating |
||
|
Preferred stock is cumulative and NON participating |
In: Finance
A 15.44 gram sample of an organic compound
containing C, H and O is analyzed by combustion analysis and
21.89 grams of CO2 and
13.44 grams of H2O are produced.
In a separate experiment, the molar mass is found to be
62.07 g/mol. Determine the empirical formula and
the molecular formula of the organic compound.
Empirical Formula?
Molecular Formula?
In: Chemistry