Answer:
software as a service(SaaS) model
Explanation:
This is because the software applications are accessed over the internet meaning no need for infrastructure installation and maintenance.
Answer:
Explanation:
Keep in mind a lossy algorithm will lose information while a lossless algorithm maintains all your original information.
Therefore:
A. False, a lossy algorithm will not allow perfect reconstruction.
B. True, if you don't care about keeping all your information it's easier to compress.
C. False, you can use a lossless algorithm for anything.
D. False, the point of lossless is that you keep all information.
Answer:
Explanation:
Based on the information provided in this scenario it can be said that this is likely due to there being a cultural lag between having the Internet and using the technology to its full capacity. Cultural lag refers to the notion that culture takes time to catch up with technological innovations, mainly due to not everyone has access to the new technology. As years pass a specific technological advancement becomes more readily accessible to the wider public as is thus more widely adopted.
<span>Large headphones are essentially just two loudspeakers mounted on a strap that clamps firmly over your head. Earbuds work
the same way but, as you would expect, everything inside them (the
magnet, the coil of wire, and the diaphragm cone that makes sound) is
shrunk down to a much smaller size.</span>
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
