Answer:
true
Explanation:
it's TRUE the correct answer is true
Answer:
Explanation:
public static int cupsToOunces (int cups) {
int ounces = cups * 8;
return ounces;
}
This is a very simple Java method that takes in the number of cups in the recipe as a parameter, converts it to ounces, and then returns the number of ounces. It is very simple since 1 cup is equal to 8 ounces, therefore it simply takes the cups and multiplies it by 8 and saves that value in an int variable called ounces.
The question above has multiple choices as below;
<span>a. </span>Wear aggregation.
<span>b.
</span>Wear mitigation.
<span>c. </span>Wear prevention
<span>d.
</span>Wear leveling
The answer is d) Wear leveling.
This technique by some SSD controllers to increase the
memory’s lifetime is called wear leveling. The mechanism for this principle is
simple: distribute the entries for all the blocks evenly so that they will wear
out evenly. Flash controller typically manages wear leveling and uses a wear
leveling algorithm to control which physical block to use.
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
