Answer:
The program is as follows:
<em>5 INPUT A,B</em>
<em>6 PROD = A * B</em>
<em>7 PRINT PROD</em>
<em>8 TOTAL = A + B</em>
<em>9 PRINT TOTAL</em>
<em>10 DIFF = A - B</em>
<em>11 PRINT DIFF</em>
<em>12 END</em>
Explanation:
This gets input for the two numbers
<em>5 INPUT A,B</em>
This calculates the product
<em>6 PROD = A * B</em>
This prints the calculated product
<em>7 PRINT PROD</em>
This calculates the sum
<em>8 TOTAL = A + B</em>
This prints the calculated sum
<em>9 PRINT TOTAL</em>
This calculates the difference
<em>10 DIFF = A - B</em>
This prints the calculated difference
<em>11 PRINT DIFF</em>
This ends the program
<em>12 END</em>
Answer:
The answer is "5.5 and 90.90%"
Explanation:
For point 1:
To calculate the speed from fast mode, its run time without enhancement should be worked out. Designers are aware of which two selves are implicated throughout the accelerated project planning: the empty (
) and the increased stages (
).
Although not enhanced, this would take almost as long (
) and for the combine to give phase; even so, the increased phase would've been 10 times longer, as well as
. Thus the corresponding total speed without enhancement is 
Its overall speed is:

For point 2:
We re-connect these figures in Amdahl's Law throughout order to identify how long it would take for both the initial implementation:


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
The answer for Apex College and Career Prep 2 would be Probable College.