Answer:
The correct answer to this question is option (1).
Explanation:
Let x1 = Is number of pages that printer A can print in page/minute.
thus x1 + 5 = Is number of page that printer B can print in page/minute.
As printer A finishes his work in 60 min. So the task has 60x1 pages. As printers A and B both finish that task in 24 minutes. So the equation can be given as:
60x1 = 24x1 + 24(x1 + 5)
60x1 = 24x1 + 24x1 + 120
60x1=48x1 + 120
60x1-48x1=120
12x1=120
x1=120/12
x1=10
put the value of x1 in to (60x1). Therefore, the task has 60(10) = 600 pages. the task contains 600 pages.
So the answer to this question is option (1).
Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line = 
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
B. hand gestures
these are not verbal.