A dental hygienist is interested in the number of cavities teenagers have when they visit the dentist. The dental hygienist believes the average number of cavities is more than 3 cavities and would like to test this claim. During the process of hypothesis testing, the dental hygienist computes a value based on the significance level and test type. This value then creates a rejection region. What value did the dental hygienist compute

A value that is higher than 3 as the average number of cavities teenagers have

Step-by-step explanation:

The value computed by the dentist which creates a rejection region means that the value falls within the range of the range of values that leads the researcher to reject the null hypothesis.

In this study, the null hypothesis is that the average number of cavities is less than or equal to three.

Thus, since the dentist's computed value falls within the rejection region, it means his computed value will be more than 3 as the average number of cavities.

Hope that helps...

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v