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3 years ago

How to do an experiment without a bias?

1 answer:

3 0

You do not have any personal connections to the people or animals in the experiment. Also people cannot know if they were given a placebo or not not because of the Placebo effect. The placebo effect is when a person knows they were given a placebo but acts a different way to give the experiment “better” results. You also should eliminate any other pre-existing bias about people you have like a person is nicer if they have a dog. This is called hindsight bias.

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if the highest isoline on a hill is 24 meters, and the interval between isolines is 3 meters, what would the next possible highe

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3 years ago

Susan was cleaning up after her party, and she had to dispose of ten items. How many will likely end up in a landfill?

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Answer:

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3 years ago

Shannon has been a member of the newspaper club for 2 years and attends writing shops in her free time. Which career choice shou

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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

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3 years ago

What does the quote from Trotsky's Diary in Exile reveal about his and his like-minded comrades' point of view regarding the fat

Nostrana [21]

Answer: It revealed the other exact political landscape and sociocultural landmark of writer's philosophical line of thoughts. It opens a wide range of iconic point of views.

Explanation: It is a varying attempt to point out the anarchical systems

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3 years ago