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iris [78.8K]
3 years ago
12

For each entity, select the attribute that could be the unique identifier of each entity.

Computers and Technology
1 answer:
Alika [10]3 years ago
3 0

Answer:

"student ID, number and title" is the correct answer for the above question.

Explanation:

  • The primary key is used to read the data of a table. It can not be duplicated or null for any records.
  • The "student ID" is used to identify the data of the "student" entity. It is because the "student ID" can not be duplicated or null for any records. It is because every student has a unique "student ID".
  • The "number" is used to identify the data of the "Locker" entity. It is because the "number" can not be null or duplicate for any records. It is because every locker has a unique "number".
  • The "title" is used to identify the data of the "Movie" entity. It is because the "title" can not be Null or duplicate for any records. It is because every movie has a unique "title".
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Three healthcare firms jointly own and share the same cloud resources to meet their computing needs. Which cloud model does this
Stels [109]

Answer:

yes

Explanation:

yall share the same thing

5 0
2 years ago
The set of coordinating colors applied to backgrounds, objects, and text in a presentation is called:
Sonja [21]

Answer:

theme colors

Explanation:

As said, a group of colors that are used to format text and objects in a document. When you open the Color menu, these colors determine what you see.

3 0
2 years ago
A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0
3 years ago
Computers that are close to one another are connected to form a LAN
Archy [21]

Explanation:

different computer are connected to a LAN by a cable and an interface card

6 0
3 years ago
Kevin is working in the Tasks folder of his Outlook account. Part of his computer screen is shown below.
klasskru [66]

Answer:

D

Explanation:

7 0
3 years ago
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