Answer:
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Step-by-step explanation:
Total plants = 11
Domestic plants = 7
Outside the US plants = 4
Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Here we have n=4, p=4/11 and q=7/11
P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰
= 1 - 0.16399
P(X≥1) = 0.836
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
*looks at question for a while* .....
OH !
If you lived in Great Britain and you wanted to buy one Euro it would cost LESS THAN A POUND .
Answer:
4, 9v, and 6w
Step-by-step explanation:
I'm pretty sure it's right
Answer:
A) None
Step-by-step explanation:
1) 
shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but 
is not a factor of nst = 144.
2) 
shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but 
is not a factor of nst = 12*24. In fact, it is a greater number.
3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.
So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.
Answer:
3 quarts
Step-by-step explanation:
divide the volume value by 4
