For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.
Then, we must find the roots of:
Where:
We have to:
Substituting we have:
By definition we have to:
So:
Thus, we have two roots:
Answer:
It's the first one because the other three go pasted -6 which would make the problem not true.
Answer: 1/5, 1/2, 0.
Step-by-step explanation:
given data:
no of cameras = 6
no of cameras defective = 3
no of cameras selected = 2
Let p(t):=P(X=t)
p(2)=m/n,
m=binomial(3,2)=3!/2!= 3
n=binomial(6,2)=6!/2!/4! = 15
p(3)= 3/15
= 1/5.
p(1)=m/n,
m=binomial(6,1)*binomial(2,2)=6!/1!/4!*2!/2!/0!= 7.5
n=binomial(6,2)= 15
p(2)= 7.5/15
= 1/2
p(0)=m/n,
m=0
p(0)=0
Answer:
45q + 35
Step-by-step explanation:
