Disk access is way slower than
memory access. Caching is a technique to improve the disk access time. Block cache is a caching technique which
reduces disk accesses time. Here a
portion of disk is bought to cache for reading and a modified blocks are first
changed in cache but reflected in the disk at one go. On the other hand with write through caching
each modified block is written to cache and at the same time it is written to
the disk. Write- through caching
requires more disk I/O so they can have a negative effect on the performance.
Answer:
The correct answers are: (A) +/- 5V and (B) +/- 12V
Explanation:
First at all, the purpose of the power supply is transforming from AC (Alternating Current) to DC (Direct Current) for use this energy in chips and electronics devices that consume this type of power. Usually the most common DC voltage for electronics is 5V but in your CPU you also have other devices that might going to need more voltage like fan or hard drives thats ´s why 12V is an also an option. Now some other devices are going to need negative sources such -5V or -12V as well.
Answer: inattention, lack of knowledge, and negligence.
Explanation: employees are often the cause of workplace incidents because they may be unknowledgeable about security protocols, negligent, or simply make a mistake.
Answer:
Answered below
Explanation:
class TestScores {
double test1;
double test2;
double test 3;
public TestScores (double test1, double test2, double teat3){
this.test1= test1;
this.test2 = test2;
this.test3 = teat3;}
//mutator
public setTest1(double test1){
this.test1 = test1;
}
//accessor
public double getTest1(){
return test 1;
}
//Write same accessors and mutators for test2 and test3
public double getTestAverage(){
double sum = test1+test2+test3;
return sum / 3;
}
}
class TestRun{
public static void main (String [] args){
TestScores scores = new TestScores(50.5, 40.0, 80.7)
int average = scores.getTestAverage();
System.out.print(average);
}
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
