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3 years ago

For each of the descriptions below, perform the following tasks:

Computers and Technology

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

attached below in the explanation and the diagrams attached

Explanation:

question

i) identify the degree and cardinality of the relationship

ii) Express the entities and the relationships with attributes in each description graphically with an E-R diagram 

A)

i) The degree and cardinality of the relationship

The entities PIANO, MODEL and DESIGNER have a degree of two i.e. Binary relationship
The cardinality of the entities PIANO, MODEL and DESIGNER have a relationship of ONE-to-many

ii) description graphically with an E-R diagram

The E-R diagrams show that The degree of the relationship for the entities PIANO, MODEL, and DESIGNER is two(2). and also shows that there is a One-to-Many cardinality available in the entities PIANO, MODEL, and DESIGNER

E-R diagram Attached below

B)

i) The degree and cardinality of the relationship

entities PIANO and Technician have a degree of two ( 2 ) ; i.e. a Binary relationship
The cardinality of the entities PIANO, and TECHNICIAN have a relationship of Many-to-Many.

ii) Graphical description with an E-R diagram

The diagram describes/ shows that, entities PIANO and Technician have a degree of two ( 2 ) ; i.e. a Binary relationship and also The cardinality of the entities PIANO, and TECHNICIAN have a relationship of Many-to-Many.

E-R diagram attached below

C)

i) Degree and cardinality of the entity Technician

The entity TECHNICIAN has a degree one, i.e. a unary relationship.

The cardinality of the entity TECHNICIAN have the relationship of One-to-Many.

ii) E-R diagram attached below

D) 

i) Degree and cardinality of the relationship

The entities TABLET TYPE and PROCESSOR TYPE have a degree of two(2) , i.e. a binary relationship

The cardinality these entities TABLET TYPE, and PROCESSOR TYPE have is a Many-to-Many relationship

ii) E-R diagram attached for TABLET TYPE, PROCESSOR TYPE is attached below

E)

i) Degree and cardinality of the relationship

The entities TABLET TYPE, TABLET COMPUTER, and CUSTOMER have a degree of two(2) ; i.e. a binary relationship.

The cardinality these entities TABLET TYPE, TABLET COMPUTER, and CUSTOMER have is a One-to-Many relationship

also the shipment to the customer is multiple hence the relationship can be said to be Many-to-Many relationship

also The attribute Shipping Date will become an attribute of that M: M relationship.

ii) The E-R diagram for TABLET TYPE, TABLET COMPUTER, CUSTOMER is attached below

F)

i) Degree and cardinality of the relationship between TABLET TYPE, TECHNICIAN

The entities TABLET TYPE, and TECHNICIAN have a degree of two(2); i.e. a binary relationship.

The cardinality these entities TABLET TYPE, and TECHNICIAN have is a Many-to-Many relationship

ii) E-R diagram attached below

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Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent server

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Answer:

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Explanation:

Given

$Reliability = 99.98\%$

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Required

Minimum number of servers needed

Let p represent the probability that a server is reliable and the probability that it wont be reliable be represented with q

Such that

$p = 95\%\\$

It should be noted that probabilities always add up to 1;

So,

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Subtract p from both sides

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To get an expression for one server

The probabilities of 1 servers having 99.98% reliability is as follows;

$p = 99.98\%$

Recall that probabilities always add up to 1;

So,

$p + q = 1$

Subtract q from both sides

$p + q - q = 1 - q$

$p = 1 - q$

So,

$p = 1 - q = 99.98\%$

$1 - q = 99.98\%$

Let the number of servers be represented with n

The above expression becomes

$1 - q^n = 99.98\%$

Convert percent to fraction

$1 - q^n = \frac{9998}{10000}$

Convert fraction to decimal

$1 - q^n = 0.9998$

Add $q^n$ to both sides

$1 - q^n + q^n= 0.9998 + q^n$

$1 = 0.9998 + q^n$

Subtract 0.9998 from both sides

$1 - 0.9998 = 0.9998 - 0.9998 + q^n$

$1 - 0.9998 = q^n$

$0.0002 = q^n$

Recall that $q = 0.05$

So, the expression becomes

$0.0002 = 0.05^n$

Take Log of both sides

$Log(0.0002) = Log(0.05^n)$

From laws of logarithm $Loga^b = bLoga$

So,

$Log(0.0002) = Log(0.05^n)$ becomes

$Log(0.0002) = nLog(0.05)$

Divide both sides by $Log0.05$

$\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}$

$\frac{Log(0.0002)}{Log(0.05)} = n$

$n = \frac{Log(0.0002)}{Log(0.05)}$

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