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2 years ago

Identify 3 reasons why birth rates fall in countries entering stage 3 and 4 of the DTM.

Advanced Placement (AP)

1 answer:

kolezko [41]2 years ago

4 0

Answer:

Explanation:

STAGE 3 - death rates are low and birth rates decrease, usually as a result of improved economic conditions, an increase in women's status and education, and access to contraception

STAGE 4 - birth rates and death rates are both low, stabilizing total population growth. ... Additionally, the decline in birth rate and death rate isn't always consistent, nor are both always in decline

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2 years ago

How and why did states in Africa develop and change over time?<br>For; AP World History: Modern

ratelena [41]

Answer:

Explanation:

Between about 800 and 1200 AD, partly in response to the new currents of trade, Africans in the southern sahel developed a set of trading states with an elaborate organization.Well before 1500 AD, a number of states had come into existence, that had no connection at all with the trade across the Sahara.

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3 years ago

How might you explain the different reasons why you may forget the process of classical conditioning if you were tested on it to

jonny [76]

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2 years ago

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antiseptic1488 [7]

Answer:

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Explanation:

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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

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2 years ago