The answer is B) Exponentially, shows a constant percentage in sales per month
The experimental research has two key advantages over correlational studies.
1. The possibility of random assignment
2. Causal connections can be assumed.
<h3>
What benefit does experimental research have over correlational research?</h3>
Correlational studies merely examine the data that already exists, whereas experimental studies give the researcher the opportunity to influence the study's factors. Researchers can make inferences about how changes in one variable affect changes in another through the use of experimental investigations.
The factors in a correlation study are out of the control of the researcher or research team. The researcher merely measures the information she discovers in the outside world. She can then determine whether changes in one are related to changes in the other, or if the two variables are correlated. In such a study, experimenters gather existing data and use statistical methods to examine it, such as economic statistics from governments. The findings of correlation studies can lead to hypotheses that can be verified through a more focused experimental one.
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Its y=3/4x+3
you can solve it
-3x+4y=12
add 3x both sides
4y=3x+12
divide 4 by both sides
y=3/4x+4
Answer:
![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Step-by-step explanation:
A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".
We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).
If we select a value
we want this:

And we can express this like that:
for each possible i
We assume that the random variable
are independent and
from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

And then cumulative distribution would be expressed like this:



For each value
we can find the dendity function like this:

So then we have the pdf defined, and given by:
and 0 for other case
And now we can find the expected value for the random variable X like this:

![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Answer: 6,300 m^3
Step-by-step explanation:
Put numbers into calculator