Answer: final notions on frequency, reach, and consistency.
If my answers wrong please tell me :D
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Answer:
Option d) B is 1.33 times faster than A
Given:
Clock time, 
No. of cycles per instructions, 
Solution:
Let I be the no. of instructions for the program.
CPU clock cycle, 
= 2.0 I
CPU clock cycle, 
= 1.0 I
Now,
CPU time for each can be calculated as:
CPU time, T = 
Thus B is faster than A
Now,
Performance of B is 1.33 times that of A
<h2>Answer and Explanation:</h2>
The picture shows the right careers with their respective career clusters.
Answer:
# Code in Python
dictionary={'A':1,'B':2,'C':3,'D':4}
other_dictionary={}
for keys in dictionary:
if dictionary[keys]&1==1:
temp=dictionary[keys]*dictionary[keys]-10*10
other_dictionary[keys]=temp
else:
other_dictionary[keys]=dictionary[keys]
print(other_dictionary)
assert other_dictionary
Explanation:
- Initialize a sample example dictionary and other_dictionary.
- Do a binary comparision for checking odd number
.
- Update the the value stored in the dictionary to store the squared difference of the original value and '10'.
- For even: store the original value (from dictionary).
