Answer:
1. Uncompressed audio formats
2. Formats with lossless compression
3. Formats with lossy compression
Explanation:
1. Uncompressed audio formats, such as WAV, AIFF, AU or raw header-less PCM;
2. Formats with lossless compression, such as FLAC, Monkey's Audio (filename extension .ape), WavPack (filename extension .wv), TTA, ATRAC Advanced Lossless, ALAC (filename extension .m4a), MPEG-4 SLS, MPEG-4 ALS, MPEG-4 DST, Windows Media Audio Lossless (WMA Lossless), and Shorten (SHN).
3. Formats with lossy compression, such as Opus, MP3, Vorbis, Musepack, AAC, ATRAC and Windows Media Audio Lossy (WMA lossy).
*This question seems like an opinion based question, but here is my opinion on the matter:
Yes the cell phone is changing our views of what is polite and impolite. It is indeed impolite to talk on your phone while you are ordering a drink at Starbucks, because then you aren't focusing on ordering instead you are focused on your phone call which can interfere with your ordering.
Hope this helps!
Answer:
The resultants bits for two strings are 0111 1110
Explanation:
Given details:
two pair of strings are
10010100
11101010
comparing two pair of strings, then perform XOR operation
Rules to follow for XOR
If both the bits are same then resultant is zero 0 and if bits are different resultant is one
XOR operation:
10010100
11 1 01010
-------------
01111110
The resultants bits for two strings are 0111 1110
def is_list_even(my_list):
for i in my_list:
if(i%2 != 0):
return False
return True
def is_list_odd(my_list):
for i in my_list:
if(i%2 == 0):
return False
return True
def main():
n = int(input())
lst = []
for i in range(n):
lst.append(int(input()))
if(is_list_even(lst)):
print('all even')
elif(is_list_odd(lst)):
print('all odd')
else:
print('not even or odd')
if __name__ == '__main__':
main()
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