The users, groups, logins, and roles that have access to a server are called ________________. a. ids b. permissions c. principals d. securables

Answers

Answer 1
Answer: The users, groups, logins, and roles that have access to a server are called securables. Correct answer: D
A securable is anything that can have permissions granted, denied, or revoked on in SQL Server, while permission is used to grant a user account access to a database.


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Given the following structure and variable definitions, struct customer { char lastName[ 15 ]; char firstName[ 15 ]; unsigned int customerNumber; struct { char phoneNumber[ 11 ]; char address[ 50 ]; char city[ 15 ]; char state[ 3 ]; char zipCode[ 6 ]; } personal; } customerRecord, *customerPtr; customerPtr = &customerRecord; write an expression that can be used to access the structure members in each of the following parts: a) Member lastName of structure customerRecord. b) Member lastName of the structure pointed to by customerPtr. c) Member firstName of structure customerRecord. d) Member firstName of the structure pointed to by customerPtr. e) Member customerNumber of structure customerRecord. f) Member customerNumber of the structure pointed to by customerPtr. g) Member phoneNumber of member personal of structure customerRecord. h) Member phoneNumber of member personal of the structure pointed to by customerPtr. i) Member address of member personal of structure customerRecord. j) Member address of member personal of the structure pointed to by customerPtr. k) Member city of member personal of structure customerRecord. l) Member city of member personal of the structure pointed to by customerPtr.

Answers

Answer:

see explaination

Explanation:

a)

customerRecord.lastName

b)

customerPtr->lastName or (*customerPtr).lastName

c)

customerRecord.firstName

d)

customerPtr->firstName or (*customerPtr).firstName

e)

customerRecord.customerNumber

f)

customerPtr->customerNumber or (*customerPtr).customerNumber

g)

customerRecord.personal.phoneNumber

h)

customerPtr->personal.phoneNumber or (*customerPtr).personal.phoneNumber

i)

customerRecord.personal.address

j)

customerPtr->personal.address or (*customerPtr).personal.address

k)

customerRecord.personal.city

l)

customerPtr->personal.city or (*customerPtr).personal.city

m)

customerRecord.personal.state

n)

customerPtr->personal.state or (*customerPtr).personal.state

o)

customerRecord.personal.zipCode

p)

customerPtr->personal.zipCode or (*customerPtr).personal.zipCode

PLEASE HURRY!!!
Look at the picture below and please help.

Answers

2 < 3 and 5 < 1 is false because 5 is not less than 1.

3 < 3 or 1 <= 1 is true because 1 is less than or equal to 1.

Not (2 = 3) is true because the opposite of 2 = 3 is true

A teacher at your school is using her district issued laptop to create spreadsheets for her part-time job as a bookkeeper. Which standard, if any, is this teacher violating?

Answers

The teacher is violating the Standard 1.2 in the education section.

What did the Standard 1.2 states?

The standard states that any educator shall not knowingly misappropriate or use monies, personnel, property, or equipment committed to his or her charger for personal gain or advantage.

Hence, because of this, the teacher has violated the Standard 1.2 in the education section.

Read more about Standard 1.2

brainly.com/question/17277092

#SPJ2

Answer:

The teacher is most likely violating the District Use Policy.

6 things you should consider when planning a PowerPoint Presentation.

Answers

Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.

Explanation:

Susan is taking a French class in college and has been asked to create a publication for her class. What feature can sheuse to help her develop her publication in French?
Research
Grammar
Language
Spell Check

Answers

the answer is Language

Answer:

Essayons

Explanation:

3. Problem 5. A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. What is the maximum allowable size for memory? d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Answers

The number of bits that are needed for this operation code (opcode) is 8 bits.

What is binary encoding?

Binary encoding can be defined as a standard technique that is designed and developed for converting data in plain text (source alphabets) to a form that is easily used by different operating systems (OS), especially through the use of a binary digit (bit) or two-symbol system.

How to calculate the number of bits.

The number of bits that are needed for this operation code (opcode) is given by this mathematical expression:

2^n > 150\n\n2^8 > 150\n\n256 > 150

n = 8 bits.

b. To determine the number of bits that are left for the address part of the instruction:

  • Opcode, O, I = 8 bits.
  • Instruction size = 24 bits.

Address = I-O\n\nAddress =24-8

Address = 16 bits.

c. To determine the maximum allowable size for memory:

Based on the memory unit, this is given by 2^(16)* 24

d. The largest unsigned binary number that can be accommodated in one word of memory is given by: 2^(24)-1

Read more on binary encoding here: brainly.com/question/5381889

Answer:

a) 8 bits b) 16 bits. c) 2¹⁶ * 24 bits d) 2²⁴ -1

Explanation:

a) In order to be able to accommodate 150 different instructions, the number of bits needed must be equal to the minimum power of 2 that satisfies this equation:

2n > 150, i.e. n=8.  

b) If the total number of bits for a word is 24, and 8 are used for the op code, there are 16 bits left for the address part.

c) If the address part has 16 bits, this means that the total addressable space is just 2¹⁶, so the maximum allowable size is 2¹⁶ * 24 bits.

d) As we have 24 bits to be filled, the largest unsigned binary number is just 2²⁴ – 1.

(As we need to leave a position for all zeros).

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