Consider a two-level cache hierarchy with L1 and L2 caches. An application incurs 1.4 memory accesses per instruction on average. For this application, the miss rate of L1 cache is 0.1; the L2 cache experiences, on average, 7 misses per 1000 instructions. The miss rate of L2 expressed correct to two decimal places is ________.

Let the total number of instructions executed be $N = 1000$.

1. Total Memory Accesses to L1:

   $$\text{Total L1 Accesses} = 1.4 \times N = 1.4 \times 1000 = 1400$$

2. Number of L1 Misses (which become accesses to L2):

   $$\text{L1 Misses} = \text{Miss Rate of L1} \times \text{Total L1 Accesses} = 0.1 \times 1400 = 140$$

3. Number of L2 Misses: Given as 7 misses per 1000 instructions, so for $N = 1000$:

   $$\text{L2 Misses} = 7$$

4. Local Miss Rate of L2 Cache:

   $$\text{Miss Rate of L2} = \frac{\text{Number of L2 Misses}}{\text{Number of L2 Accesses}} = \frac{7}{140} = \frac{1}{20} = 0.05$$

Thus, the miss rate of L2 is 0.05.