X=10
-3 on both sides
Then -6(x+4)+3 to the other side
P(most favorable outcome) = 1 -(0.03 +0.16 -0.01) = 0.82
_____
"repair fails" includes the "infection and failure" case, as does "infection". By adding the probability of "repair fails" and "infection", we count the "infection and failure" case twice. So, we have to subtract the probability of "infection and failure" from the sum of "repaire fails" and "infection" in order to count each bad outcome only once.
The probability of a good outcome is the complement of the probability of a bad outcome.
Answer:the frist on
Step-by-step explanation:
there 5 canby bars and there are 8 boxes so you would do(5x8) divided by s and s= 1
<h3>Sample space = {a,b,c,d,e,f}</h3><h3>Event space = {a,c}</h3>
We simply list all of the letters mentioned as they are the possible outcomes. We can only pick one item from the sample space. The event space is the set of outcomes where we want to happen (picking either an 'a' or 'c').
Step-by-step explanation:
(5√2-4√3)(5√2-4√3)=
(5√2-4√3)^2=(5√2)^2+(4√3)^2-2(5√2)(4√3)
=25*2+16*3-10√2*4√3
=50+48-10√2*4√3
=98-10√2*4√3 is the answer