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Answer:
B. Analyze data from the car’s external cameras.
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Answer:
A) Number of bits for byte = 6 bits
B) number of bits for index = 17 bits
C) number of bits for tag = 15 bits
Explanation:
Given data :
cache size = 64 kB
block size = 32 -byte
block address = 32 -bit
number of blocks in cache memory
cache size / block size = 64 kb / 32 b = 2^11 hence the number of blocks in cache memory = 11 bits = block offset
A) Number of bits for byte
= 6 bits
B) number of bits for index
block offset + byte number
= 11 + 6 = 17 bits
c ) number of bits for tag
= 32 - number of bits for index
= 32 - 17 = 15 bits
Answer:
2^32 times as many values can be represented
Explanation:
32-bit. This means that the number is represented by 32 separate one’s and zero’s. 32 bits of 2 possible states = 2^32=4,294,967,296 possible values.
Integer meaning that only whole multiples of one are accepted.
Signed meaning that negative values are accepted. This halves the number of possible positive values (roughly), so the largest number you can represent is 2^31–1=2,147,483,647, but instead of 0, the smallest number you can represent is -2,147,483,648. An unsigned 32-bit integer, by contrast, can represent anything from 0 to 4,294,967,295.