Making charts, files that need complicated calculations
Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:

Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:

Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:

Speedup gain is 1.81 times.
The clearer the resolution screen is it is better seen through the graphics. Say you have a bad computer screen, well your graphics are going to be bad. The better and higher quality it is will be the better you can see your images.
Solution :
a.
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b).
getTitle() {
return
;
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setTitle(
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this.
=
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getAuthor() {
return author;
}
setAuthor(String author) {
this.
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;
}
getPublisher() {
return
;
}
setPublisher(String
) {
this.
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}
public int get
() {
return
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set
(int
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this.
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}
Answer:
Some of the qualities of the installed memory which HWiNFO can reveal are:
- Date of manufacture
- Memory type
- Speed of memory
Explanation:
HWiNFO is an open-source systems information app/tool which Windows users can utilize as they best wish. It helps to detect information about the make-up of computers especially hardware.
If you used HWiNFO to diagnose the memory, the following information will be called up:
- the total size of the memory and its current performance settings
- the size of each module, its manufacturer, and model
- Date of manufacture
- Memory type
- Speed of memory
- supported burst lengths and module timings,
- write recovery time etc.
Cheers