Answer:
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
Step-by-step explanation:
Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
This is because at a higher level of discrepancies, there will be a strong evidence against the null. This means that it will be rare to find discrepancies if null was true.
In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.
Plug in f(6) so it would be -2/3(6)+5= 1
So ur answer would be one
Im assuming the problem is x^2-16, which would be factored to (x-4)(x+4)
Answer:
Step-by-step explanation:
The scale factor is given as 1:500,000
i.e. original 500,000 is represented as 1 inch in the map
Hence if 12.7 inches on the map, this is a question of direct variation
So actual distance = 12.7(500000) = 6350000
