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Aneli [31]
3 years ago
11

Define the following terms: data, database, DBMS, database system, database catalog, program-data independence, user view, DBA,

end user, canned transaction, deductive database system, persistent object, meta-data, and transaction-processing application.
Engineering
1 answer:
nalin [4]3 years ago
6 0

Answer:

data - Any form of value in a column of a table in a relational database.

DBMS - Short for Database management system, which is a software that can be used to create, manipulate and view databases, e.g. MS SQL Server.

database system - Same as DBMS.

database catalog - Place where the metadata of a Database including its tables, users etc. exists, e.g. date created, size, number of columns etc. Also known as Data Dictionary.

program-data independence - Program-data independence refers to the design of keeping different levels of database (external, logical and physical) separate and loosely coupled from each other, for easier maintenance or modification work.

DBA - Short for Database Administrator, person responsible for maintaining the database. Its main responsibility is to keep the data clean and safe i.e. data doesn't contain wrong or invalid data, and is safe from viruses and is backed up.

end user - Anyone who is not directly interacting with a database, but through some software, like website or mobile application.

persistent object - An object of a class in a program, that interacts with the underlying database and is responsible for manipulating the database, the program is connected to.

transaction-processing system - A transaction process system is part of a software, responsible for making sure critical business transactions, like crediting or debiting money, either goes through cleanly or doesn't at all.

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3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
Why are funeral home adding dining facilities to the business
Gelneren [198K]
Because they think it will make them more money
5 0
3 years ago
What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
Eight energy corporations made plans to increase their combined spending on efficiency programs to $50 million per year for the
tangare [24]

Answer:

F=531831381

Explanation:

There are two  ways of doing this question:

1) By Formula

2) By Using  Compound interest Table

By Formula:

F=A*\frac{(1+i)^{n}-1}{i}

Where:

F is future value

A is annual amount per year

i is interest rate

n is number of years

F=50 million*\frac{(1+0.08)^{8}-1}{0.08}

F=531831381

By Using  Compound interest Table:

F=A(F/A,i,n)

From Table F/A at i=8% and n=8 is 10.6366

F=50000000*(10.6366)

F=531830000≅531831381

The difference in two answers is due to decimal points if you take value from table to greater decimal points you will get the exact answer as by using formula.

7 0
3 years ago
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