37. 37-4=33 33÷3=11 hope i helped
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.
Since M divides segment AB into a ratio of 5:2, we can say that M is 5/(5+2) of the length of AB. Therefore 5/7 × AB.
distance of AB = d
5/7×(x2 - x1) for the x and 5/7×(y2 - y1) for the y
5/7×(8 - 1) = 5/7 (7) = 5 for the x
and 5/7×(16 - 2) = 5/7 (14) = 10 for the y
But remember the line AB starts at A (1, 2),
so add 1 to the x: 5+1 = 6
and add 2 to the y: 10+2 = 12
Therefore the point M lies exactly at...
A) (6, 12)
