Answer: big boi like a someboody 2x -853
Step-by-step explanation:
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.
X*y=72
x = 5y-2
x=72/y
72/y=5y-2
72=5y^2-2y
5y^2 -2y - 72 = 0
(5y+18)(y-4)=0
y = -18/5 discard
or y = 4
4x=72
x=18
(18,4)
First I found the number of adults by multiplying .60 by 700. It was 420. From there all I had to do was subtract it from 700. Revealing the number of children to be 280.
