Answer B) Headers
Explanation:
I just took a Tech class, im pretty sure its headers. Hope this helps! -3-
Answer:
import random
def grid_maker(x, y):
return [ [str(random.randint(-30, 30)) for _ in range(x)]for _ in range(y) ]
print ('\n'.join(' '.join(row) for row in grid_maker(4, 5)))
Explanation:
We use a template function and we generate numbers between -30 and 30:
>>> [str(random.randint(-30, 30))]
We also use the str() method so we can later concatenate the integers with \n. If you do not specify the integers, it will cause a crash. Also, keep in mind if we use a variable to store the integers, it will come out more of like a seed than a random grid. For instance, output without the random integers in a variable:
-12 16 -18 -3
7 5 7 10
18 -21 -16 29
21 3 -4 10
12 9 6 -9
Vs with a variable:
-25 6 -25 -20
-25 6 -25 -20
-25 6 -25 -20
-25 6 -25 -20
-25 6 -25 -20
Next we specify the x and the y:
>>> for _ in range(x)]for _ in range(y) ]
Next we just print it and create a new line every time a row is made
hope this helped :D
Answer:
The answer is B
Explanation:
12 cm, 7 cm, and 5 cm.
Add 5 + 7 and get 12
Sides added must be greater, not equal to.
Answer:
2 x 10⁵ bytes per second
Explanation:
Given:
MIPS rate = maximum speed of CPU to execute instructions = 10 million instructions per seconds
number of instructions required to transfer 1 byte using interrupt driven I/O = 50
Maximum number of bytes that can be transferred in 1 second = MIPS rate / number of instructions for 1 byte
=> max number of bytes = 10 million / 50 = 10 x 10⁶ / 50 = 2 x 10⁵
which is less than the maximum transfer rate of memory = 100 million bytes per second
So, maximum data transfer rate during I/O operations by using interrupt-driven I/O is 2 x 10⁵ bytes per second
Answer:
Merge sort is a sorting technique based on divide and conquer technique.
Explanation:
MERGE(A, p, q, r)
n1 = q - p + 1
n2 = r - q
L[1..n1] and R[1..n2] this creates the new array
for i = 1 to n1
L[i] = A[p + i - 1]
for j = 1 to n2
R[j] = A[q + j]
i = 1
j = 1
for k = p to r
if i > n1
A[k] = R[j]
j = j + 1
else if j > n2
A[k] = L[i]
i = i + 1
else if L[i] ≤ R[j]
A[k] = L[i]
i = i + 1
else
A[k] = R[j]
j = j + 1
