Answer:
Given Data:
Clock rate of P1 = 4 GHz
Clock rate of P2 = 3 GHz
Average CPI of P1 = 0.9
Number of Instructions = 5.0E9 = 5 × 10^9
Clock rate of P2 = 3 GHz
Average CPI of P2 = 0.75
Number of Instructions = 1.0E9 = 10^9
To find: If the computer with largest clock rate has the largest performance?
Explanation:
Solution:
As given in the question, clock rate of P1 = 4 GHz which is greater than clock rate of P2 = 3 GHz
According to the performance equation:
CPU Time = instruction count * average cycles per instruction/ clock rate
CPU Time = I * CPI / clock rate
Where instruction count refers to the number of instructions.
Performance of P1:
CPU Time (P1) = 5 * 10^9 * 0.9 / (4 * 10^9)
= 5000000000 * 0.9 / 4000000000
= 4500000000 / 4000000000
= 1.125s
Performance of P2:
CPU Time (P2) = 10^9 * 0.75/ (3 * 10^9)
= 750000000 / 3000000000
= 0.25s
So the Performance of P2 is larger than that of P1,
performance (P2) > performance (P1)
0.25 is better than 1.125
But clock rate of P1 was larger than P2
clock rate of P1 > clock rate of P2
4 GHz > 3 GHz
So this is a misconception about P1 and P2.
It is not true that computer with the largest clock rate as having the largest performance.
Answer:
Internal sound card
Explanation:
In this case, George has used an Internal sound card, this is enough to record a podcast, make and receive video conferences or play video games.
If we're going to use an instrument, an external sound card, it's necessary, but in this case, George can make the record with high quality audio.
If George wants to add an adapter to connect 6'35mm but is not the same quality that an external sound card.
Answer:
See the table attached for complete solution to the problem.
Answer:
65
Explanation:
Answer:
65
Explanation:
Scroll Lock
Pause/Break
Backspace
Answer:
D. Refrigerants
Explanation:
In the United States of America, the agency which was established by US Congress and saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, pesticide use, contamination, and hazardous waste spills is the Environmental Protection Agency (EPA). Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act .
The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants such as R-12 and R-134a. This ban became effective on the 1st of January, 1993.
Refrigerants refers to any chemical substance that undergoes a phase change (liquid and gas) so as to enable the cooling and freezing of materials. They are typically used in air conditioners, refrigerators, water dispensers, etc.
The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants. The correct option is D.
Thus, The Environmental Protection Agency (EPA), which was formed by US Congress, is the organization tasked with regulating all facets of pollution, environmental cleanup, pesticide use, contamination, and hazardous material spills.
All refrigerants, including R-12 and R-134a, are forbidden from being released during service under the Clean Air Act Amendments of 1990. On January 1st, 1993, this ban came into force.
Any chemical compound that transforms into a different phase (liquid or gas) to allow for the cooling or freezing of items is referred to as a refrigerant. They are frequently found in refrigerators, water dispensers, air conditioners, and other appliances.
Thus, The Clean Air Act Amendments of 1990 prohibit service-related releases of all refrigerants. The correct option is D.
Learn more about Clean air act, refer to the link:
#SPJ3
Explanation:
A binary number is converted to hexadecimal number by making a group of 4 bits from the binary number given.Start making the group from Least Significant Bit (LSB) and start making group of 4 to Most Significant (MSB) or move in outward direction.If the there are less than 4 bits in last group then add corresponding zeroes to the front and find the corresponding hexadecimal number according to the 4 bits.For example:-
for this binary number 100111001011011.
100 11100101 1011
There is one bits less in the last group so add 1 zero(0) to it.
0100 1110 0101 1011
4 E 5 B
100111001011011 = 4E5B
101011.1010
0010 1011 . 1010
2 B A
=2B.A