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ser-zykov [4K]
2 years ago
5

Which of the following statements is true? A. Interpreted programs run faster than compiled programs. B. Compilers translate hig

h-level language programs into machine language programs. C. Interpreter programs typically use machine language as input. D. None of the above.
Computers and Technology
2 answers:
alina1380 [7]2 years ago
6 0

Answer:

B.

Explanation:

Computer programs are written in high level languages. Every language has its own set of syntax and semantics.

Computer only understands instructions in machine language which is written using strings of 0 and 1.

Thus, for execution of a high level language program, it is mandatory to convert high level instructions into machine instructions.

This translation into machine language is done in two ways – compilation or interpretation.

Some languages are compiled while others are interpreted.

Both compilers and interpreters are programs and process the high level language program as needed.

Compiler

Any program written in high level language is compiled.

The program is first analyzed before compilation. This process takes time. The analysis phase is called pre processing.

The program compilation stops only when the program is completely analyzed. Hence, it is difficult to locate the error in the program. This makes debugging easy.

Debugging is defined as removal of errors in the programs.

A compiler is responsible for converting high level language code into machine level code.

The complete program undergoes compilation at the same time.

A compiled program executes faster than an interpreted program.

A compiled program produce machine code which is a collection of 0 and 1, hence, it takes less time to execute.

Examples of compiled languages are C and C++.

Interpreter

Some high level languages are interpreted.

The high level language code is translated into an intermediate level code. This intermediate code is then executed.

The analysis of the program takes less time and takes place before interpretation begins. The analysis phase is called pre processing.

An interpreted program is translated one line at a time.

The program interpretation stops when an error is encountered. This makes debugging easy.

Debugging is defined as removal of errors in the programs.

Examples of interpreted languages include Python and Ruby.

denpristay [2]2 years ago
6 0

Answer:

The correct answer is B. Compilers translate high-level language programs into machine language programs.

Explanation:

Compilers translate high-level language programs into machine language programs, It is a program that translates the human-readable code to an object/machine code in other to translate the source code into an executable program.

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tensa zangetsu [6.8K]

Answer: d) Exploit

Explanation: Exploit is a type computer attack that successful when the computer system of an user is vulnerable and attacker can do the exploitation. This happens due to the weakness of the system, applications software, network etc.

Other given option are incorrect because exit door,glitch and bad are not any type of attack in the computer field that causes harm to the system.Thus the correct option is option(d).

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The answers are 1, 3, and 5.
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2 years ago
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9. Which of the following are true for all regular languages and all homomorphisms? (a) h (L1 ∪ L2)= h (L1) ∩ h (L2). (b) h (L1
mixer [17]

Answer:

h (L1 ∪ L2)= h (L1) ∩ h (L2).

This is true.

There will be w ∈  (L1 ∪ L2) for any s ∈ h (L1 ∪ L2) in such a way that s=h(w)

we can assume that w ∈ L1

So In this case h(w) ∈ L (S1). Hence s ∈ L(S1)

for any s ∈ h (L1) U h(L2)

We can assume that s ∈ L(S1)

There exists w ∈ L1 such that s= h(w)

In this case it is w ∈ L1 U L2 as well.

Hence , s ∈ h (L1 U L2)

Explanation

consider  = 0,1 and  = a,b and h(0) = a , h(1) = ab

(a) Consider L1 = 10,01 and L2 = 00,11

   Now L1 ∪ L2 = 00,01,10,11

   h (L1 ∪ L2) = h(00) , h(01) , h(10) , h(11) = h(0)h(0) , h(0)h(1) , h(1)h(0) , h(1)h(1)

  = aa, aab , aba , abab

Hence h (L1 ∪ L2) = aa, aab , aba , abab .

Here h (L1) = h(10) , h(01) = h(1)h(0) , h(0)h(1) = aba , aab

Hence h (L1) = aba , aab .

Here h (L2) = h(00) , h(11) = h(0)h(0), h(1)h(1) = aa, abab

Hence h(L2) = aa, abab.

Finally Hence , h (L1 ∪ L2)= h (L1) ∩ h (L2).

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Answer:

see explaination

Explanation:

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Customer:

The scheme allows the customer to do below mention actions:

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To create version to be able to purchase invention from the structure.

Browse crop through search category or shortest search alternative.

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Make sum for the order(s) positioned.

Payment System:

Payment system allows client to pay using subsequent two methods:

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System allow seller to perform underneath actions:

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Create account to happen to a member.

Administrator:

Following actions are perform by Admin:

Manage the goods posted in the system.

Modify existing manufactured goods categories or add novel categories for foodstuffs.

Site Manager:

System privileges Site director with the following role in the system:

View information on:

Orders Placed by customer

Products added by seller

Accounts created by users

check attachment

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Answer:

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# see comment for the rest. Brainly won't let me save an answer which contains code.

Explanation:

Above is your program fixed, but it is unclear on how to deal with unprintable characters, i.e., characters that originally were near the highest printable character ('~')

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