Answer:
Answer is b
Step-by-step explanation:
cuh cuh cuh cuh cuh cuh , , , , , ,,
Answer:
c= 0 m=6
Step-by-step explanation:
0x2= 0
3x6=18
18+0=18$
<span>0.3046% chance of damaging at least one of five payloads.
First, determine how many standard deviations it takes for the parachute to open at 100 meters. So
200 m - 100 m = 100 m
100 m / 31 m = 3.2258
Looking up on a standard normal table, the probability of having that deviation or higher is 0.00061, or 0.061%
Since you're looking for the probability of damaging at least one of five, the easiest way of calculating that is to subtract from 1 the probability of having all five parachutes successfully deploy. So
1 - (1 - 0.00061)^5 = 1 - 0.99939^5 = 1 - 0.996953719 = 0.003046281 = 0.3046%</span>
Answer:
It doesn't make sense, i can solve this problem
Because 9 has to go into each number, if it doesn't, then it can't be a factor, nor great common factor
