You have a network with a subnet of 172.16.17.0/22. Which of the following is a valid host address?A. 172.16.17.1 255.255.255.25
1 answer:
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Answer:
Explanation:
The following code is written in Python. It is a function called checkString that takes in a string as an argument and loops through each char in that string and checking to see if it is lowercase, uppercase, vowel, consonant, or punctuations. It adds 1 to the correct variable. At the end of the loop it prints out all of the variables. The picture below shows a test output with the string "Brainly, Question."
def checkString(word):
uppercase = 0
lowercase = 0
vowel = 0
consonants = 0
punctuation = 0
vowelArray = ['a', 'e', 'i', 'o','u', 'y' ]
consonantArray = ['b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z']
punctuationArray = ['.', '!', '?', ',', ';', '(', ')']
for char in word:
if char.isupper():
uppercase += 1
else:
lowercase += 1
if char.lower() in vowelArray:
vowel += 1
elif char.lower() in consonantArray:
consonants += 1
if char in punctuationArray:
punctuation += 1
print('Uppercase: ' + str(uppercase))
print('Lowercase: ' + str(lowercase))
print('Vowels: ' + str(vowel))
print('Consonants: ' + str(consonants))
print('Punctuations: ' + str(punctuation))
Answer:
Check the explanation
Explanation:
CPI means Clock cycle per Instruction
given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M
Execution time is given by following Formula.
Execution Time(CPU time) = CPI*Instruction Count * clock time =
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is
= 333.33 micro sec
c) Since the system B is slower than system A, So the system A executes the given program in less time
Hence take CPU execution time of system B as CPU time of System A.
therefore
216.67 micro = =
Instructions = 216.67*750/2.5
= 65001
hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A