how can you hack on a cumputer witch one chrome hp

Answer:

10.b

11.c

12.c

13.a

14.d

15.b

16.c

Explanation:

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Answer:

Following are the responses to these question:

Explanation:

The cache size is 2n words whenever the address bit number is n then So, because cache size is 216 words, its number of address bits required for that  cache is 16 because the recollection is relational 2, there is 2 type for each set. Its cache has 32 blocks, so overall sets are as follows:

[tex]\text{Total Number of sets raluired}= \frac{\text{Number of blocks}}{Associativity}[/tex]

                                               [tex]=\frac{32}{2}\\\\ =16\\\\= 2^4 \ sets[/tex]

The set bits required also are 4. Therefore.

Every other block has 8 words, 23 words, so the field of the word requires 3 bits.

For both the tag field, the remaining portion bits are essential. The bytes in the tag field are calculated as follows:

Bits number in the field tag =Address Bits Total number-Set bits number number-Number of bits of words

=16-4-3

= 9 bit

The number of bits inside the individual fields is therefore as follows:

Tag field: 8 bits Tag field

Fieldset: 4 bits

Field Word:3 bits