Answer:
Fault-Tolerance
Explanation:
Fault tolerance refers to the ability of a system (computer, network, cloud cluster, etc.) to continue operating without interruption when one or more of its components fail.
Answer:
e) Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".
Explanation:
<em>if - elseif - else statements work in sequence in which they are written. </em>
- <em> </em>In case <em>if() statement is true, </em>else if() and else statements will not get executed.
- In case <em>else if() statement is true</em>, conditions in if() and else if() will be checked and else statement will not be executed.
- In case <em>if() and else if() both are false</em>, else statement will be executed<em>.</em>
First, let us consider code segment I.
In this, first of all "pea" is checked in if() statement which will look for "pea" only in the String str. So, even if "pearl" or "pear" or "pea" is present in 'str' the result will be true and "pea" will get printed always.
After that there are else if() and else statements which will not get executed because if() statement was already true. As a result else if() and else statements will be skipped.
Now, let us consider code segment II.
In this, "pearl" is checked in if() condition, so it will result in desired output.
Executable code is attached hereby.
Correct option is (e).
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark">
java
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<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark">
java
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Answer:
c.
Explanation:
I believe that in this scenario, the best option for this data would be a hash table using open addressing with 1,800 entries. Hash tables consume more memory than lists but it makes up for it with much faster response time speeds. This is because hash tables work on a key:value system therefore, the license plate can easily be grabbed from the database extremely quickly by just plugging in the plate number. Doing so will retrieve all of the saved information from that license plate. That is why hash tables have a constant time complexity of O(1)
Answer:
CPU need 50% much faster
disk need 100% much faster
Explanation:
given data
workload spend time CPU = 60%
workload spend time I/O = 40%
achieve overall system speedup = 25%
to find out
How much faster does CPU need and How much faster does the disk need
solution
we apply here Amdahl’s law for the overall speed of a computer that is express as
S = 
.............................1
here f is fraction of work i.e 0.6 and S is overall speed i.e 100% + 25% = 125 % and k is speed up of component
so put all value in equation 1 we get
S =
1.25 = 
solve we get
k = 1.5
so we can say CPU need 50% much faster
and
when f = 0.4 and S = 125 %
put the value in equation 1
S =
1.25 = 
solve we get
k = 2
so here disk need 100% much faster
Answer:
public class GetInfo{
Beverage[] beverages=new Beverage[100];
int i=0;
GetInfo(Beverage b){
beverages[i]=b;
i++;
}
public void Display(){
for(int i=0;i<beverages.length;i++)
cout<<beverage[i].tostring();
}
Explanation:
we are taking Beverages array to store all values and in constructor we are adding that to the list and Display() function prints the vale
