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Aneli [31]
4 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]4 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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What is the chord length of an airplane called?
aliya0001 [1]

Answer:

The distance from the leading edge to the trailing edge is called the chord, denoted by the symbol c. The ends of the wing are called the wing tips, and the distance from one wing tip to the other is called the span, given the symbol s

3 0
3 years ago
Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest gen
Anvisha [2.4K]

Answer:

a) Sample size = 1691

b) 95% Confidence Interval = (0.3696, 0.4304)

Explanation:

(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?

The margin of error is given by

MoE = z \cdot \frac{\sqrt{p(1-p)} }{\sqrt{n} }  \\\\

Where z is the corresponding z-score for 90% confidence level

z = 1.645 (from z-table)

for p = 0.50 and 2% margin of error, the required sample size would be

n = \frac{1.645^{2} \cdot 0.50(1-0.50)}{0.02^{2}}  \\\\n = \frac{0.6765}{0.0004}  \\\\n = 1691\\

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find  a 95% confidence interval for p.

The sample proportion is

p = 400/1000

p = 0.40

z = 1.96 (from z-table)

n = 1000

The confidence interval is given by

CI = p \pm z \cdot \sqrt{\frac{p(1-p)}{n} } \\\\CI = 0.40 \pm 1.96 \cdot \sqrt{\frac{0.40(1-0.40)}{1000} } \\\\CI = 0.40 \pm 1.96 \cdot 0.01549 \\\\CI = 0.40 \pm 0.0304 \\\\CI = 0.40 - 0.0304 \: and \: 0.40 + 0.0304\\\\CI = (0.3696 ,\:  0.4304)

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

3 0
3 years ago
State three means of operating a power tool
Shtirlitz [24]
Power tools are usually run on one of the three types of power: Compressed air, electricity, or combustion
3 0
2 years ago
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in
Mama L [17]

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

d_{p} =mN_{p}

Where

m = module = 5

Np = number of teeth of pinion = 26

d_{p} =5*26=130mm = 0.13 m

The pitch line velocity is equal to:

V_{t} =\frac{d_{p}*2*\pi  *w_{p} }{120}

Where

wp = speed of the pinion = 1800 rpm

V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s

The factor B is equal to:

B=\frac{(12-Q_{v})^{2/3}  }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}

The bending stress is equal to:

\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}}  \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa

4 0
3 years ago
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