Yes, they're equivalent values.
Answer:
79239729379249
Step-by-step explanation:
your aswer is 284389300 oo
Answer:
Let X be the number of times the target is hit. The probability P(X≥1) then equals 1 minus the probability of missing the target three times:
P(X≥1) = 1− (1−P(A)) (1−P(B)) (1−P(C))
= 1−0.4*0.3*0.2
= 0.976
To find the probability P(X≥2) of hitting the target at least twice, you can consider two cases: either two people hit the target and one does not, or all people hit the target. We find:
P(X≥2)=(0.4*0.7*0.8)+(0.6*0.3*0.8)+(0.6*0.7*0.2)+(0.6*0.7*0.8) = 0.788
Step-by-step explanation:
I think it is $15.12
I hope that helps!!!!!!
: )
