Answer:
A) <em><u>a,s,d,f, and either g or caps lock</u></em>
B) <u><em>l,k,j,h, and, ;</em></u>
Explanation:
Answer:
There is also an attachment below
Explanation:
Since we are talking about binary search, let's assume that the items are sorted according to some criteria.
Time complexity of binary search is O(logN) in worst case, best case and average case as well. That means it can search for an item in Log N time where N is size of the input. Here problem talks about the item not getting found. So, this is a worst case scenario. Even in this case, binary search runs in O(logN) time.
N = 700000000.
So, number of comparisions can be log(N) = 29.3 = 29.
So, in the worst case it does comparisions 29 times
Answer:
a) the average CPI for machine M1 = 1.6
the average CPI for machine M2 = 2.5
b) M1 implementation is faster.
c) the clock cycles required for both processors.52.6*10^6.
Explanation:
(a)
The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4
= 1.6
The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4
= 2.5
(b)
The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6
Given 80MHz = 80 * 10^6
The average MIPS ratings for M1 = 80 x 10^6 / 1.6 x 10^6
= 50
Given 100MHz = 100 * 10^6
The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6
= 40
c)
Machine M2 has a smaller MIPS rating
Changing instruction set A from 2 to 1
The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)
and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.
Answer:
a is the correct answer
Explanation:
correct me if I'm wrong hope it's help thanks
Explanation:
Explanation:
RSA encryption is performed by calculating C=M^e(mod n).
However, if n is much larger than e (as is the case here), and if the message is not too long (i.e. small M), then M^e(mod n) == M^e and therefore M can be found by calculating the e-th root of C.
