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
Answer:
a. 255.255.255.0 (class C)
b. 255.255.255.224
Explanation:
Here, we want to give the implied subnet mask of the given classful IPV4 address
We proceed as follows;
Given IPv4 address: 200.200.200.200
Classes
Class A : 0.0.0.0 to 127.255.255.255
Class B: 128.0.0.0 to 191.255.255.255
Class C: 192.0.0.0 to 223.255.255.255
Class D: 224.0.0.0 to 239.255.255.255
so 200.200.200.200 belongs to Class C and it's subnet mask is 255.255.255.0
In CIDR(Classless Inter Domain Routing)
subnet /27 bits means 27 1s and 5 0s. so subnet
i.e 11111111.11111111.11111111.11100000 which in dotted decimal format is 255.255.255.224 .
Answer:
The sum of all positive even values in arr
Explanation:
We have an array named arr holding int values
Inside the method mystery:
Two variables s1 and s2 are initialized as 0
A for loop is created iterating through the arr array. Inside the loop:
num is set to the ith position of the arr (num will hold the each value in arr)
Then, we have an if statement that checks if num is greater than 0 (if it is positive number) and if num mod 2 is equal to 0 (if it is an even number). If these conditions are satisfied, num will be added to the s1 (cumulative sum). If num is less than 0 (if it is a negative number), num will be added to the s2 (cumulative sum).
When the loop is done, the value of s1 and s2 is printed.
As you can see, s1 holds the sum of positive even values in the arr
Answer:
for y in range(88, 43, -4):
print(y, end=" ")
Explanation:
yw
Answer:
Software requirement specification
Explanation:
requirement specification is the first step before proceeding with any project.
we should validate our output at all phases with software requirement specification
