Answer:
Option a: corrective action
Explanation:
A corrective action is a way to offer the remedy to a process or product which fail to meet expectation and doesn't work as intended. This is one of the management aspect to improve the quality of a product or process. By running corrective action, the undesirable damage brought by the erroneous or faulty process can be eliminated or at least reduce to certain extend.
Answer:
A) 2.56 ms
B) 128 μs
Explanation:
Time for Bus cycles = 500 ns
Transfer of bus control = 250 ns
I/0 device data transfer rate = 50 KB/s
Data are transferred at : 1 byte at a time
A) Determine how long the device tie up the bus when transferring a block of 128 bytes
Block size to be transferred = 128 bytes
Bandwidth = 50 KB/s
data transfer = (Block size ) / ( Bandwidth)
= ( 128 * 8 ) / ( 50 * 10^3 * 8 )
= ( 1024 ) / ( 50 * 10^3 * 8 )
= 2.56 ms
To determine the actual transfer time we have to add up the transfer time for the bus control in both directions :
2.56 ms + 500 ns = 2.56 ms. this is because 500 ns is not a significant value
B) cycle stealing mode
In this mode each byte is transferred at a time and The total transfer time needed for the bus control in both directions will be double the total time i.e 2 * 500 ns = 1000 ns. because additional control time of 250 ns is required at both ends
since 1 byte is transferred at a time , 1 byte will be transferred in 1 μs
128 bytes = 128 * 1 μs = 128 μs
Answer:
PROGRAM QuadraticEquation
Solver
IMPLICIT NONE
REAL :: a, b, c
;
REA :: d
;
REAL :: root1, root2
;
//read in the coefficients a, b and c
READ(*,*) a, b, c
WRITE(*,*) 'a = ', a
WRITE(*,*) 'b = ', b
WRITE(*,*) 'c = ', c
WRITE(*,*)
// computing the square root of discriminant d
d = b*b - 4.0*a*c
IF (d >= 0.0) THEN //checking if it is solvable?
d = SQRT(d)
root1 = (-b + d)/(2.0*a) // first root
root2 = (-b - d)/(2.0*a) // second root
WRITE(*,*) 'Roots are ', root1, ' and ', root2
ELSE //complex roots
WRITE(*,*) 'There is no real roots!'
WRITE(*,*) 'Discriminant = ', d
END IF
END PROGRAM QuadraticEquationSolver
False, the government CAN control the internet, but thankfully it is not owned by them. Anybody with access to the internet can run it.
Calculate the sum of the first 5 positive odd integers: Let's do this in our head first, so we can check if our code is right or not!
The first positive 5 odd integers are: 1, 3, 5, 7, 9
Sum these to: 25
int sum = 0, k; <------These just declare our variables, telling the program 'Hey, I'm going to use 'sum' and 'k' to store data.
for (k = 1; <---We're going to repeat the following code, and we're starting at 1
k <= 10; <---- We're going to continue to repeat until we greater than 10.
k += 2) <------ Every time we do a loop, we're going to add 2.
{ sum += k; } <---- We're going to increase the number inside "sum" by the number inside "k"
Let's run this and see what happens. Remember, we keep going until we hit more than 10.
Round 0: k = nothing, sum = 0 (before we start the loop)
Round 1: k = 1, sum = 1
Round 2: k = 3, sum = 1+3 or 4
Round 3: k = 5, sum = 4 + 5 or 9
Round 4: k = 7, sum = 9 + 7 = 16
Round 5: k = 9, sum = 16 + 9 = 25
Round 6: k = 11, sum = 25 + 11 = 36
Well, we can tell here that round 5 was correct, but round 6 is not correct. And our loop condition says <=10, which means we have to do Round 6.
This means we did it one too many times. Our ending condition should be <10, instead of <=10.
Option B
