Press and hold down the "Fn" button while pressing the "Home Prt Sc" button above the numeric keypad to capture the entire screen. Make a screenshot of the active window only by holding down the "Fn" and Alt" keys while pressing "Home Prt Sc." 2. Press the "Start" button, and type "Paint" into the search box.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
First, we need to convert these hexadecimal numbers into decimal numbers, then we can easily identify which one is the lowest hexadecimal.
The hexadecimal numbers are F2, 81, 3C, and 39.
F2 = (F2)₁₆ = (15 × 16¹) + (2 × 16⁰) = (242)₁₀
81 = (81)₁₆ = (8 × 16¹) + (1 × 16⁰) = (129)₁₀
3C = (3C)₁₆ = (3 × 16¹) + (12 × 16⁰) = (60)₁₀
39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀
The 39 is the lowest hexadecimal number among the given numbers.
Because 39 hex is equal to 57 decimal.
39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀
Answer:
Option B is the correct answer.
Explanation:
- In the above code, the loop will execute only one time because the loop condition is false and it is the Do-While loop and the property of the Do-while loop is to execute on a single time if the loop condition is false.
- Then the statement "x*=20;" will execute one and gives the result 200 for x variable because this statement means "x=x*20".
- SO the 200 is the answer for the X variable which is described above and it is stated from option B. Hence it is the correct option while the other is not because--
- Option A states that the value is 10 but the value is 200.
- Option C states that this is an infinite loop but the loop is executed one time.
- Option D states that the loop will not be executed but the loop is executed one time
There isn't an opening and closing parentheses for the else statement. PM me. I know a lot of batch...
Answer:
Explanation:
The following Python program uses a combination of dictionary, list, regex, and loops to accomplish what was requested. The function takes a file name as input, reads the file, and saves the individual words in a list. Then it loops through the list, adding each word into a dictionary with the number of times it appears. If the word is already in the dictionary it adds 1 to its count value. The program was tested with a file named great_expectations.txt and the output can be seen below.
import re
def wordCount(fileName):
file = open(fileName, 'r')
wordList = file.read().lower()
wordList = re.split('\s', wordList)
wordDict = {}
for word in wordList:
if word in wordDict:
wordDict[word] = wordDict.get(word) + 1
else:
wordDict[word] = 1
print(wordDict)
wordCount('great_expectations.txt')
