Answer:
Bounce between different objects
Explanation:
Light probably will bounce off of objects or it might go through, i think it depends on what object and what material.
1. ms-dos 1981
2. windows 2.0 1987
3. windows xp 2001
4. mac os 2001
5. windows 7 2009
Alternative 1:A small D-cache with a hit rate of 94% and a hit access time of 1 cycle (assume that no additional cycles on top of the baseline CPI are added to the execution on a cache hit in this case).Alternative 2: A larger D-cache with a hit rate of 98% and the hit access time of 2 cycles (assume that every memory instruction that hits into the cache adds one additional cycle on top of the baseline CPI). a)[10%] Estimate the CPI metric for both of these designs and determine which of these two designsprovides better performance. Explain your answers!CPI = # Cycles / # InsnLet X = # InsnCPI = # Cycles / XAlternative 1:# Cycles = 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150)CPI= 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150) / X1= X(0.50*2 + 0.50(0.94*2 + 0.06*150) ) / X= 0.50*2 + 0.50(0.94*2 + 0.06*150)= 6.44Alternative 2:# Cycles = 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150)CPI= 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150) / X2= X(0.50*2 + 0.50(0.98*(2+1) + 0.02*150)) / X= 0.50*2 + 0.50(0.98*(2+1) + 0.02*150)= 3.97Alternative 2 has a lower CPI, therefore Alternative 2 provides better performance.
Answer:
The solution code is written in Python 3:
- def modifyList(listNumber):
- posCount = 0
- negCount = 0
-
- for x in listNumber:
- if x > 0:
- posCount += 1
- else:
- negCount += 1
-
- if(posCount == len(listNumber)):
- listNumber.append(max(listNumber))
-
- if(negCount == len(listNumber)):
- listNumber.append(min(listNumber))
-
- print(listNumber)
-
- modifyList([-1,-99,-81])
- modifyList([1,99,8])
- modifyList([-1,99,-81])
Explanation:
The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).
To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).
If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).
If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.
At last we can print the listNumber (Line 17).
If we test our function using the three sets of input list, we shall get the following results:
[-1, -99, -81, -99]
[1, 99, 8, 99]
[-1, 99, -81]
The Rudolph Rule states that simple ways you can make information stand out and guide or satisfy your audience to important details and highlight important information in your presentation
so i conclude option D is correct for above statement
hope it helps
