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rusak2 [61]
4 years ago
6

Why is demand for fresh strawberries elastic?

Advanced Placement (AP)
1 answer:
OLga [1]4 years ago
6 0

Answer:

The easy access to substitutes for berries is likely to be one reason why demand is own-price elastic. Among the four types of berries, the retail demand for strawberries is the least responsive to the price with an elasticity of-1.26.

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A psychologist conducted a study at her home during an annual activity of children wearing masks and going door-to-door receivin
Zarrin [17]

The graph is attached.

Answer:

a) Dependent variable is the variablr that is measured by the psychologist.

Here the dependent variable can be said to be the number of children who collected additional candy.

b) The hypothesis 1 and hypothesis 2 here are:

H1 : The children would take more candy when they were alone.

H2: The children would take more candy when they were masked.

From the data given(graph attached), the data does not support the hypothesis 1, because from the graph the percentage of children taking additional candy is higher when they are in group.

The given data supports the hypothesis 2 which states that the children would take more candy when they were masked.

c) The psychologist cannot generalize her findings to all children because she only used children in her neighborhood. This can be called sampling bias.

d) This study is not a naturalistic behaviour because the psychologist is manipulating a variable. She may choose to increase or decrease the number of masked or unmasked children.

e)

i) Modeling: In this case, the children may take additional candy because they saw others doing so.

ii) Deindividuation: The chances of the children taking additional candy when they were in group is more likely, because they felt anonymous. That's why the percentage of children who took extra candy was higher when they were in group.

iii) Lawrence Kohlberg’s preconventional stage:

In this case a child may not take additional candy because of avoidance of punishment.

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3 years ago
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Is planting a vegetable garden a procedural or declarative knowledge?
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Procedural (procedural is the knowledge of how to DO something, declarative is more like facts and is conscious learning)
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2 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
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