You basiclly divide 6 and 1.2 and get your answer.
Answer:
Independent variable: amount of water
Dependent variable: growth of the plant
Step-by-step explanation:
In the statement "Julie notices that her plant grows one inch for every liter of water that it receives" it is implied that the growth of the plant is related proportionally to the amount of water it receives.
We know the amount of growth in function of the amount of water
The dependant variable, the result, is the growth of the plant.
Then, the independent variable is the amount of water, as it is the input to calculate the amount of growth.
I believe about two thirds
Answer:
16
Step-by-step explanation:
10+6=16
Answer:
Binomial
There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.
Step-by-step explanation:
For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
In this problem
Of the 20 copies, 2 are defective, so
.
What is the probability that you will encounter neither of the defective copies among the 10 you examine?
This is P(X = 0) when
.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.1%29%5E%7B0%7D.%280.9%29%5E%7B10%7D%20%3D%200.3487)
There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.