7×3=21 so he has 7 more lawns or 2 1/3 months
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.
Answer:
A. 370.58
Step-by-step explanation:
Hope that helps
Answer:
12
Step-by-step explanation:
12/2=6
replace x with 12
Well 0.65 as a percentage is 65% which is 65/100. So in the simplest for it would be the highest number going into both which would be 5 and then divide both numbers by 5.
65/5=13
100/5=20
So the answer for this question would be 13/20
