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devlian [24]
3 years ago
12

Dim Inventory() as Integer = {357, 126, 220}Call Stock(Inventory(2),Me.lblOutput)Sub Stock (ByVal Number As Integer, ByRef lblLa

bel As Label)Me.lblLabel.Text = NumberEnd SubWhich is displayed in the label when the code is executed?a) 357b) 126c) 220d) None of the above
Computers and Technology
1 answer:
aalyn [17]3 years ago
6 0

Answer:

The answer is "Option c".

Explanation:

In the given visual basic code, an integer array "Inventory" is declared, that includes 3 elements, that are "357, 126, and 220". In the next line, a call stock function is used, in the function, the array is passed as a parameter and in array parameter, the index value that is "2" is passed. This function use label to print array index value, that is "220". and other options are incorrect that can be described as follows:

  • Option a and Option b both are wrong because array indexing always starts with 0 and ends with n-1, and in this array, it will take index value 2, which is not equal to 357 and 126.

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Write a Python program called wdcount.py which uses a dictionary to count the number of occurrences of each word (ignoring case)
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The program is an illustration of loops.

<h3>What are loops?</h3>

Loops are program statements used to perform repetition

<h3>The wordcount.py program</h3>

The program written in Python, where comments are used to explain each line is as follows

# This opens the file in read mode

text = open("myFile.txt", "r")

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2 years ago
Given the following code segment, how can you best describe its behavior? i ← 1 FOR EACH x IN list { REMOVE(list, i) random ← RA
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Question is attached as image, please help :&gt;
harkovskaia [24]

Answer:

Proved

Explanation:

Given

X = (a.\bar b)+(\bar a.b)

(a + b)\ .  \frac{}{a.b}

Required

Find out why they represent the same

To do this, we simplify either (a + b)\ .  \frac{}{a.b} or X = (a.\bar b)+(\bar a.b)

In this question, I will simplify (a + b)\ .  \frac{}{a.b}

Apply de morgan's law

(a + b)\ .  \frac{}{a.b} = (a + b) . (\bar a + \bar b)

Apply distribution property

(a + b)\ .  \frac{}{a.b} = a.\bar a + a.\bar b + \bar a . b + b . \bar b

To simplify, we apply the following rules:

a.\bar a = 0 --- Inverse law

b.\bar b = 0 --- Inverse law

So, the expression becomes

(a + b)\ .  \frac{}{a.b} = 0 + a.\bar b + \bar a.b + 0

(a + b)\ .  \frac{}{a.b}  = a.\bar b + \bar a.b

Rewrite as:

(a + b)\ .  \frac{}{a.b}  = (a.\bar b)+(\bar a.b)

From the given parameters:

X = (a.\bar b)+(\bar a.b)

This implies that:

(a + b)\ .  \frac{}{a.b} when simplified is X or (a.\bar b)+(\bar a.b)

3 0
3 years ago
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