20 = 2 x 2 x 5
15 = 3 x 5
common divisor (CD) of 20 and 15 is 5 so n should be 5 and each group has 4 boys and 3 girls.
If there are more than one CD, then there are more than one answer.
If they ask for the largest number of groups, greatest common divisor (GCD) will be the answer.
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.
Y=(3x+4)(-2x-3)=-6x^2-4x-9x-12=-6x^2-13x-12.
It is a quadratic function because it has x in power 2 (x^2)
18 tens= 180
20 ones= 20
180+20= 200
200
0.54=0.54/1=5.4/10=54/100=27/50
