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Elis [28]
3 years ago
8

Which of the following is a valid variable name? a. salesTax b. input-string c. 25Percent d. double

Computers and Technology
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

The correct answer of the given question is option(A) i.e salesTax

Explanation:

Variable are the storage area which value varies during the execution of aprogram .

some of the rules of variable are

1 Variable always start with alphabet.

2 We cannot give variable name any reserve keyword.

3 Variable cannot start with digit.

4 We cannot use special symbol like -,$ to declare any variable

In option (b ) input-string ,it is invalid variable because it uses special symbol "-"  to declared the variable.

In option (c) 25Percent , it is invalid variable because it violate the rule of  variable and start with digit.variable cannot start with digit

In option( d )double, it is invalid variable because it violate the rule of  variable because double is keyword and variable never be a keyword.

So option(A) is correct it follow the rule of variable.

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. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege
Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) \Omega is transitive, which means it converts one action to others object because if \Omega(f(n))=g(n) indicates c.g(n). It's true by definition, that becomes valid. But if \Omega(g(n))=h(n), which implies c.h(n). it's a very essential component. If c.h(n) < = g(n) = f(n) \. They  \Omega(f(n))   will also be h(n).  

In point (ii), The  value of \Theta is convergent since the \Theta(g(n))=f(n). It means they should be dual a and b constant variable, therefore a.g(n) could only be valid for the constant variable, that is  \frac{1}{a}\ \  and\ \ \frac{1}{b}.

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of  A is a polynomial-time complexity, which is similar to its outcome that is O(n^{2}). It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop(n^{2}). All internal loops operate on a total number of N^{2} generations and therefore the final time complexity is O(n^{2}).

6 0
3 years ago
Write a program whose input is a string which contains a character and a phrase, and whose output indicates the number of times
otez555 [7]

Answer:

The program written in python is as follows

def countchr(phrase, char):

     count = 0

     for i in range(len(phrase)):

           if phrase[i] == char:

                 count = count + 1

     return count

phrase = input("Enter a Phrase: ")

char = input("Enter a character: ")

print("Occurence: ",countchr(phrase,char))

Explanation:

To answer this question, I made use of function

This line defines function countchr

def countchr(phrase, char):

This line initializes count to 0

     count = 0

This line iterates through each character of input phrase

     for i in range(len(phrase)):

This line checks if current character equals input character

           if phrase[i] == char:

The count variable is incremented, if the above condition is true

                 count = count + 1

The total number of occurrence is returned using this line

     return count

The main method starts here; This line prompts user for phrase

phrase = input("Enter a Phrase: ")

This line prompts user for a character

char = input("Enter a character: ")

This line prints the number of occurrence of the input charcater in the input phrase

print("Occurence: ",countchr(phrase,char))

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3 years ago
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Answer:

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Explanation:

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Answer:

The answer is ".pcapng"

Explanation:

Wireshark format includes a "dump" with data packets, which is collected over a channel and sometimes a common folder to store, that data in the PCAP Next Iteration file system.

  • The .pcapng stands for the file system, this file system compatible with the recorded data transmission packet. It includes several data blocks.
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