Answer:
2500 kb
Explanation:
Here, we are to calculate the bandwidth delay product
From the question, we are given that
band width = 500 Mbps
The bandwidth-delay product is = 500 x 10^6 x 25 x 10^-3
= 2500 Kbits
Answer:
The solution is as follows.
class LFilters implements Lock {
int[] lvl;
int[] vic;
public LFilters(int n, int l) {
lvl = new int[max(n-l+1,0)];
vic = new int[max(n-l+1,0)];
for (int i = 0; i < n-l+1; i++) {
lvl[i] = 0;
}
}
public void lock() {
int me = ThreadID.get();
for (int i = 1; i < n-l+1; i++) { // attempt level i
lvl[me] = i;
vic[i] = me;
// rotate while conflicts exist
int above = l+1;
while (above > l && vic[i] == me) {
above = 0;
for (int k = 0; k < n; k++) {
if (lvl[k] >= i) above++;
}
}
}
}
public void unlock() {
int me = ThreadID.get();
lvl[me] = 0;
}
}
Explanation:
The code is presented above in which the a class is formed which has two variables, lvl and vic. It performs the operation of lock as indicated above.
Answer:
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D. Time
It’s easy to access any time even after it was just purchased.
I hope this helped
Answer:
Ig it'd be - I caught a disease when i was on a holiday.