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Dimas [21]
4 years ago
12

A case competitions database:You work for a firm that has decided to sponsor case competitions between teams of college business

students, and you were put in charge of creating a database to keep the corresponding data. The firm plans to hold about dozens of different regional competitions at various dates that will take place at different branches around the country. For each competition, you need to store a name, date, and the name of the branch where it is going to take place. For each college that has agreed to participate in the competition, you need to store the college name, a contact phone number, and a contact address. Each college can participate in only one competition, and the database should know which competition each college is participating in. On the other hand, each college is allowed to send more than one team to its competition. Each team gives itself a name and a color, and consists of several students of the same college; for each student, you want to store a first name, last name, date of birth, major, and expected graduation date. Question: How many tables do you need
Computers and Technology
1 answer:
Strike441 [17]4 years ago
3 0

Answer:

We will ned (4) four tables.

Explanation:

For the given scenario we will have to build a relational database. The database will have four tables i.e. Competation, Collage, Team and student.

For each table the database fields are mention below. Note that foreign keys are mentioned in itallic.

Competition:

Competition_ID, Competition_Name, Competition_Data, Competition_Name _of_Branch, <em>College_ID</em>

<em />

Collage:

Collage_ID, Collage_Name, Collage_Contact, Collage_address

Team:

Name, Color, <em>Team_ID, College_ID, Competition_ID</em>

<em />

College:

Student_ID, First_Name, Last_Name, Date_of_Birth, Major, Expected_Gradiuation_Date, <em>Collage_ID, Team_ID, Competation_ID</em>

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Identify the selector in the following CSS code:
SashulF [63]

Answer:

The right answer is: Option A. h3

Explanation:

The purpose of using selectors in CSS is to find the elements in the HTML page to which the formatting will be applied.

Different type of selectors are used in CSS.

In the given code, h3 is the selector.

All the <h3> elements on the page will be of orange color and will have font size 12.

Hence,

The right answer is: Option A. h3

6 0
3 years ago
Which sentence best describes the executive branches of state
melamori03 [73]

Answer:

B... I think...

Explanation:

The executive branch of every state is headed by an elected Governor. Most states have a plural executive, in which several key members of the executive branch are directly elected by the people and serve alongside the governor.

8 0
3 years ago
All it services and servers are equally critical. <br> a. True <br> b. False
g100num [7]

False

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4 0
3 years ago
Do you believe that OOP should be phased out and we should start working on some alternative(s)?
katen-ka-za [31]

Answer:

The answer to this question can be defined as below:

Explanation:

There are many new directions for solutions for any given problem in today's computer age, in which the "OOP's" is one of the philosophies of preference.  

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5 0
4 years ago
The memory capacity in bits for performing the operation y = f (x) using the table lookup method, where x is an 8-bit number and
Flura [38]

Answer:

The page field is 8-bit wide, then the page size is 256 bytes.

Using the subdivision above, the first level page table points to 1024 2nd  level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.

First, the stack, data and code segments are at addresses that require having 3 page tables entries active in the first level page table. For 64K, you need 256 pages, or 4 third-level page tables. For 600K, you need 2400 pages, or 38 third-level page tables and for 48K you need 192 pages or 3 third-level page tables. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 3 * 2 (3 second-level page tables) + 64 * (38+4+3)* 2 (38 third-level page tables for data segment, 4 for stack and 3 for code segment) = 9344 bytes.

Explanation:

16 E the answer

5 0
3 years ago
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