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levacccp [35]
3 years ago
8

Describing each stage in the demographic transition model using CBR, CDR, NIR. Characterize the amount of growth of each stage (

low, high, aka moderate).
Advanced Placement (AP)
1 answer:
SIZIF [17.4K]3 years ago
8 0
Stage 1
     Low Growth
<span>     Very high CBR, Very high CDR, <span>Very low NIR
Stage 2
     
High Growth
     High CBR, Rapidly declining CDR, Very high NIR
Stage 3
     Moderate Growth
     Rapidly declining CBR, Moderately declining CDR, Moderate NIR
Stage 4
     Low Growth
     Very low CBR, Low or slightly increasing CDR, zero or negative NIR</span></span>
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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
A. Using information in the graph shown, compare the data trend from 1960 to 2015 for sub-Saharan Africa to the trend shown for
expeople1 [14]

Answer:

A. From the information given in the graph, we have;

The grain yield in pounds per acre for Europe in 1960 is approximately 1,600 Pounds per Acre

The grain yield in pounds per acre for Europe in 2015 is approximately 5,100 Pounds per Acre

The rate of change of grain yield in Europe ≈ (5,100 - 1,600)/(2015 - 1960) = 63.\overline{63} Pounds per Acre per year

The grain yield in pounds per acre for Sub-Saharan Africa in 1960 is approximately 800 Pounds per Acre

The grain yield in pounds per for Sub-Saharan Africa 2015 is approximately 1,250 Pounds per Acre

The rate of change in grain yield in Sub-Saharan Africa ≈ (1,250 - 800)/(2015 - 1960) = 8.\overline{18} Pounds per Acre per year

Therefore, the grain yield per acre per year in Europe is increasing approximately 8 times the increase in Sub-Saharan Africa

B. Given that the rate of increase in the pounds of grain per Acre per year in Europe is progressively increasing above the average rate of world rate, the market for the grains produced in Europe will be partly the developing countries, thereby reducing the number of farms found in developing countries

C. In stage 4 demographic transition the population growth is stabilizing and therefore where there is an increase in grain yields, there will be an increase in farming and economic activity and a reduction in population growth

Where there is a decrease in grain yield, there will be an increase in activities to restore supply and therefore, a reduction in population

D. One limitation of the data in describing food security at a regional scale is the information required on the population figures in a given region involved in grain farming  

Explanation:

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Answer:

5/9*3/5

\frac{5}{9}  \times \frac{3}{5}

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