Juan have to print 9 photos.
<u><em>Explanation</em></u>
Suppose, Juan have to print 
number of photos.
First photo shop charges $0.36 per print. So, the total cost to print number of photos in this shop 
dollar.
Second photo shop charges $2.52 plus $0.08 per print. So, the total cost for printing number of photos 
As, the cost of printing is the same at both shops, so.....
So, Juan have to print 9 photos.
A sample is best defined as 'a smaller group selected from a given population'.
Answer:
What is meant by "bias due to selective survival" in cross-sectional studies?
First lets understand what is selective bias in cross-sectional studies. Basically its the bias that occurs when the random sample data for some analysis is selected in improperly. Such a selected sample is not able to represent the population that is to be analyzed. The reason is the improper randomization. So "bias due to selective survival" means that only the survival participants or survivors can be considered in this cross sectional studies. So if there is more probability of exposed cases to survive than unexposed cases or it could also be possible that unexposed cases have more probability to survive than exposed ones, so in either case the conclusion drawn from this cross sectional studies may differ from the proper cohort study. So the selective bias is introduced. Hence we can say that there is bias due to selective survival.
Step-by-step explanation:
Under what circumstances might there be no selective survival bias even if the selection probabilities are not all equal?
If the cross product of selection probabilities equals 1. Then there might be no selective survival bias even if the selection probabilities are not all equal. As we know that the cross product of odd ratio of selection probabilities is 1 means there is no relation between exposure and consequence or outcome and there is no bias in odd ratio.
Suppose that you could assess that the direction of possible selective survival bias in your study was towards the null. If your study data yielded a non-statistically significant odds ratio of 1.04, would it be correct to conclude that there was no exposure-disease association in your source population?
It would not be correct to conclude that there was no exposure-disease association in your source population because if the possible selective survival bias in the study was towards the null then this implies that the true odds ratio would be larger than 1.04. and therefore likely to be large and also differ statistically for the null value.
Answer:
(4 - a)^2 = 16-8a + a^2
(3x − 1)^2 = 9x^2−6x+1
Step-by-step explanation:
Multiply the left for each equation and they are the same.
