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Sidana [21]
3 years ago
6

6. An art museum owns a large volume of works of art. Each work of art is described by an item code (identifier), title, type, a

nd size; size is further composed of height, width, and weight. A work of art is developed by an artist, but the artist for some works is unknown. An artist is described by an artist ID (identifier), name, date of birth, and date of death (which is null for still living artists). Only data about artists for works currently owned by the museum are kept in the database. At any point in time, a work of art is either on display at the museum, held in storage, away from the museum as part of a traveling show, or on loan to another gallery. If on display at the museum, a work of art is also described by its location within the museum. A traveling show is described by a show ID (identifier), the city in which the show is currently appearing, and the start and end dates of the show. Many of the museum works may be part of a given show, and only active shows with at least one museum work of art need be represented in the database. Finally, another gallery is described by a gallery ID (identifier), name, and city. The museum wants to retain a complete history of loaning a work of art to other galleries, and each time a work is loaned, the museum wants to know the date the work was loaned and the date it was returned.As you develop the ERD for this problem, follow good data naming guidelines.

Computers and Technology
1 answer:
baherus [9]3 years ago
8 0

Answer:

Answer:

Finding entities in the given question:

1. Art

2. Artist

3. Museum

4. Location

5.Travelling

6.Gallery

Finding attributes of above entities:

1. Art (item ID, title, type, size). Here size further composed of height, width, and weight.

2. Artist (artist ID, name, date of birth, date of death, nationality)

3.Travelling (show ID, start date, end date)

4. Gallery (gallery ID, name, city)

Finding relationship between the entities:

1. Developed

2.Display

3. Part

4. Loan

Explanation:

In the below attached diagram, the primary keys are listed as, “Item ID”, “Artist ID”, “Show ID”, and “Gallery ID”.

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Sphinxa [80]

Answer:

Code

import csv

def open_file():

try:

fname=input("Enter the filename to open (xxx.yyy): ")

fp=open(fname,encoding='utf-8')

return fp

except FileNotFoundError:

print("File not found! Please Enter correct filename with extension")

return open_file()

def read_file(fp):

fields = []

rows = []

csvreader = list(csv.reader(fp))

# skiped 1 row because it contains field name

fields = csvreader[0]

D1={}

D2={}

D3={}

# accessing the data of csv file

for line in csvreader[1:]:

name = line[0].lower().strip()

platform = line[1].lower().strip()

if line[2]!='N/A':

year = int(line[2])

else:

year=line[2]

genre = line[3].lower().strip()

publisher = line[4].lower().strip()

na_sales = float(line[5])

eur_sales = float(line[6])

jpn_sales = float(line[7])

other_sales = float(line[8])

global_sales=(na_sales+eur_sales+jpn_sales+other_sales)*1000000

if name not in D1:

D1[name]=[]

D1[name].append((name, platform, year, genre, publisher,global_sales))

if genre not in D2:

D2[genre]=[]

D2[genre].append((genre, year, na_sales, eur_sales,jpn_sales, other_sales, global_sales))

if publisher not in D3:

D3[publisher]=[]

D3[publisher].append((publisher, name, year, na_sales,eur_sales, jpn_sales, other_sales, global_sales))

# sorting

temp={}

# sorting keys

for Keys in sorted (D1) :

ls=D1[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D1=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D2) :

ls=D2[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D2=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D3) :

ls=D3[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D3=temp.copy()

return D1,D2,D3

 

def main():

fp=open_file()

D1,D2,D3=read_file(fp)

# displaying the row_count in dictionaries

cnt=0

for d in D1:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D2:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D3:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

 

 

if __name__=="__main__":

main()

Explanation:

see code and see output attached

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The answer to the question is B.
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An important principle in information security is the concept of layers of security, which is often referred to as layered secur
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Answer:

Available options in question are

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Explanation:

Defense in depth consists of three major domains like physical, technical and administrative. Layered Security will consists of best practices by combining multiple  security methods to mitigate vulnerabilities and attack vectors. It is used to protect resources at different levels.  

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UkoKoshka [18]

Answer:

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Explanation:

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Simply stated, GDP is a measure of the total income of all individuals in an economy and the total expenses incurred on the economy's output of goods and services in a particular country. Also, Gross Domestic Products (GDP) is a measure of the production levels of any nation.

Basically, the four (4) major expenditure categories of GDP are consumption (C), investment (I), government purchases (G), and net exports (N).

Hence, the standard of living of the people living in a particular country automatically improves if a nation's level of productivity or production improves; they are able to easily pay for goods and services, as well as save and invest their money.

In contrast, inflation and high unemployment rate are indications of economic downturn, recession and low level of productivity (output) in a country; this would automatically affect the standard of living within such countries.

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2 years ago
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Answer: The FP-s are stored as sign (1.)111111111111 - number of bits.

Also 24 bits resolution there can be 23 zeros

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The –1·2²⁴ might be stored as -1 · 1.00000000000000000000000(1 ←the 2⁰), it depends on how the FFP “engine” manages this, it may also be code specific a n+1–n does return 1 but 1–n+n does not. you should carry out a test for a specific compiler/computer

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