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sammy [17]
3 years ago
6

Dr. Greene is planning to study methods to increase vocabulary levels in college students, using competition. Students will comp

lete forms to agree to participation in the study. He will use the teaching methods designed in the new program. He will then regularly test the students to see how much, if any, their vocabulary level increases. Students will compete against each other for prizes in the classroom. Students will not be allowed to stop the program once they start it. Student scores will be posted in the classroom with their names, to foster greater competition between them. What might an ethics review committee decide about the proposal? A The proposal is approved, as it meets the ethical guidelines defined by the APA. B The proposal is approved, because there are no animal subjects. C The proposal is denied, because participants will likely suffer psychological harm. D The proposal is denied, because confidentiality is not protected and participants cannot exercise informed consent. E The proposal is denied, because Dr. Green intends to misuse deception.
Advanced Placement (AP)
1 answer:
Illusion [34]3 years ago
4 0

Answer:

D. The proposal is denied, because confidentiality is not protected and participants cannot exercise informed consent.

Explanation:

D. is the correct answer because if the results and names of the students would be posted for others to see that is a vialation of their confidentiality.

I also took the test and that was to correct answer.

Hope this Helped ;P

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Slope integrals for calculus, having trouble need all the help I can get
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(1) The integral is straightforward; <em>x</em> ranges between two constants, and <em>y</em> ranges between two functions of <em>x</em> that don't intersect.

\displaystyle\int_{-2}^1\int_{-x}^{x^2+2}\mathrm dy\,\mathrm dx

(2) First find where the two curves intersect:

<em>y</em> ² - 4 = -3<em>y</em>

<em>y</em> ² + 3<em>y</em> - 4 = 0

(<em>y</em> + 4) (<em>y</em> - 1) = 0

<em>y</em> = -4, <em>y</em> = 1   →   <em>x</em> = 12, <em>x</em> = -3

That is, they intersect at the points (-3, 1) and (12, -4). Since <em>x</em> ranges between two explicit functions of <em>y</em>, you can capture the area with one integral if you integrate with respect to <em>x</em> first:

\displaystyle\int_{-4}^1\int_{y^2-4}^{-3y}\mathrm dx\,\mathrm dy

(3) No special tricks here, <em>x</em> is again bounded between two constants and <em>y</em> between two explicit functions of <em>x</em>.

\displaystyle\int_1^5\int_0^{\frac1{x^2}}\mathrm dy\,\mathrm dx

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