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Thepotemich [5.8K]
3 years ago
7

Has anyone taken AP Psych?

Advanced Placement (AP)
1 answer:
beks73 [17]3 years ago
4 0

Answer:

Not me

Explanation:

Because I’m in middle schol

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A key drawback with the use of tests to screen individuals for desirable characteristics is that
DochEvi [55]

Answer:

A key drawback with the use of tests to screen individuals for desirable characteristics is that It is a time-consuming procedure.

It is very important for the employees to know exactly what they must look into and assess the characteristics accurately. This process is very time consuming and takes up a lot of patience and resources to carry the tests.

A study that would need to occur quickly cannot indulge into this testing process because of it being very tedious and time consuming.

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3 years ago
The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0
3 years ago
while both cases deal with the topic of campaign finance, explain one element of McCutcheon v. FEC(2014) that is distinguishable
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Yeah it’s 100 per minute I think
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1. Identify a difference between League of United Latin American Citizens v. Perry and Shaw v. Reno.
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Answer:

May 2, 2019 — Identifying the overall constitutional principle ... In League of United Latin American Citizens v. Perry, the ... v. Perry, the Court ruled that the map in Texas was not unconstitutional on ... A. One difference between Shaw v Reno and the mentioned court case is that in Shaw v Reno, the court decided that the.

Explanation:

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List the 15 main bureaucratic agencies.
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