A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 95%, and the subsequent hit rate of the level 2 cache is 80%. An access to level 1 cache requires one cycle, and access to level 2 cache requires 3 cycles, and an access to main memory requires 80 cycles. What is the average access time in clock cycles?

Step-by-step explanation:

H1 = Hit rate of level 1 cache = 0.95

T1 = Access time for level 1 cache = 1

H2 = Hit rate of level 2 cache = 0.8

T2 = Access time for level 2 cache = 3

Hm = Hit rate of Main Memory = 1

Tm = Access time for Main Memory = 80

Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1-H1)(1-H2)*Hm*Tm]

= [0.95*1]+[1-0.95]*0.8*3]+[(1-0.95)(1-0.8)*1*80]

= [0.95+0.05*0.8*3]+[0.05*0.2*1.80]

= 2.4+0.018 = 2.418

Average access time = 2.418

2? if not the correct answer you might need to provide more information

step-by-step explanation:

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.