4 ms·
In that case, it would be: 0.51 * ((10+0)/2*0.32 + (100+11)/2*0.43 + (1000+101)/2*0.14 + (5000 + 1001)/2 * 0.03 + (100000+5001)/2*0.01 + (250000+100001)/2*0.
by christiangenco 15y ago
In that case, it would be:
0.51 * ((10+0)/2*0.32 + (100+11)/2*0.43 + (1000+101)/2*0.14 + (5000 + 1001)/2 * 0.03 + (100000+5001)/2*0.01 + (250000+100001)/2*0.01)
= $1,258.46