Answer:
The value of result is 20
Explanation:
Given
The above code segment
Required
The value of result
In the first line, we have:
In the second, we have:
This implies that:
<span>Random access memory.
This problem requires you to know what the different types of memory are and their relative advantages and disadvantages. Let's look at them and see why 3 are wrong and one is correct.
read-only memory: Otherwise known as ROM, this type of memory stores code that can't be over written. Used frequently for constant lookup values and boot code. Since it can't be written to by normal programs, it can't hold temporary values for Samantha. So this is the wrong choice.
random-access memory: Otherwise known as RAM, this type of memory is used to store temporary values and program code. It is quite fast to access and most the immediately required variables and program code is stored here. It can both be written to and read from. This is the correct answer.
hard disk: This is permanent long term readable and writable memory. It will retain its contents even while powered off. But accessing it is slow. Where the contents of RAM can be accessed in nanoseconds, hard disk takes milliseconds to seconds to access (millions to billions of times slower than RAM). Because it's slow, this is not the correct answer. But it's likely that Samantha will save her spreadsheet to hard disk when she's finished working with it so she can retrieve the spreadsheet later to work on again.
compact disk: This is sort of the ROM equivalent to the hard disk. The data stored on a compact disk can not be over written. One way of describing the storage on a compact disk is "Write Once, Read many times". In most cases it's even slower than the hard disk. But can be useful for archiving information or making backups of the data on your computer.</span>
Answer:
thanks for teaching us how to debug
Explanation:
Explanation:
A.)
we have two machines M1 and M2
cpi stands for clocks per instruction.
to get cpi for machine 1:
= we multiply frequencies with their corresponding M1 cycles and add everything up
50/100 x 1 = 0.5
20/100 x 2 = 0.4
30/100 x 3 = 0.9
CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8
We find CPI for machine 2
we use the same formula we used for 1 above
50/100 x 2 = 1
20/100 x 3 = 0.6
30/100 x 4 = 1.2
CPI for m2 = 1 + 0.6 + 1.2 = 2.8
B.)
CPU execution time for m1 and m2
this is calculated by using the formula;
I * CPI/clock cycle time
execution time for A:
= I * 1.8/60X10⁶
= I x 30 nsec
execution time b:
I x 2.8/80x10⁶
= I x 35 nsec
Usually a graphics card would be plugged in manually through a PCI-e express slot on the motherboard. But to give the card enough power (based on how powerful the card is) you would use a 6 pin or an 8 pin power connector.
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