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N76 [4]
1 year ago
9

TQ Automation What is Robotic Process Automation (RPA)?

Computers and Technology
1 answer:
tamaranim1 [39]1 year ago
5 0

Answer:

RPA or Robotic Process Automation allows organizations to automate tasks which human beings were doing across any applications and systems. The purpose of RPA is to transfer the execution of the process from humans to robots.

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: Show that the propositions (p ∨ q) ∧ (¬p ∨ r) and (p ∧ r) ⊕ (¬p ∧ q) are logically equivalent.
Ipatiy [6.2K]

Answer:

(p\vee q)\wedge(\neg p\vee r)\equiv(p\wedge r)\vee(\neg p\wedge q)

Explanation:

The statements are logically equivalent if they have the same truth tables. So let´s use truth tables in order to determine if they are logically equivalent or not:

The picture that I attached you shows the truth table for each case. As you can see in the highlight columns:

(p\vee q)\wedge(\neg p\vee r)\equiv(p\wedge r)\vee(\neg p\wedge q)

They are  logically equivalent because they have exactly the same truth values between each other. Hence, we can conclude that they are logically equivalent.

3 0
3 years ago
What are the two components that make up the chipset?
svet-max [94.6K]
Northbridge and Southbridge components.
6 0
3 years ago
A digital computer has a memory unit with 16 bits per word. The instruction set consists of 72 different operations. All instruc
OLga [1]

Answer:

a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1

Explanation:

a. How many bits are needed for the opcode?

Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72

72 = 64 + 8 = 2⁶ + 8

Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128

So, we require 7 bits for the opcode.

b. How many bits are left for the address part of the instruction?

bits left = bits per word - opcode bit = 16 - 7 = 9 bits

c. What is the maximum allowable size for memory?

Since there are going to be 2⁹ bits to addresses each word and 16 bits  for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.

We convert this to bytes

8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB

d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus

2¹⁶ - 1

6 0
3 years ago
Finish this for brainliest and 30 points
n200080 [17]
What do you mean what are we supposed to finish

6 0
3 years ago
Consider the following two code segments: Segment 1: while (k > 0) { System.out.println(k); k--; } --------------- Segment 2:
zepelin [54]

Answer:

Always same value

Explanation:

These both code segments will always generate identical values given that we provide same value of k to both segments.

You can see attached pictures.

6 0
3 years ago
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