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WARRIOR [948]
3 years ago
12

Global population is expected to increase by three billion or more over the next 50–75 years, and the number of people living in

urban areas will more than double. Most of the world's population growth will occur in developing countries. More than 1 billion people do not have access to safe and affordable drinking water. In fact, inadequate drinking water quality is a leading cause of infant mortality worldwide. What proposed solution may help with this problem?
A) increased storage facilities for clean water
B) decrease the number of motor vehicles sold in the United States
C) development of newer, faster methods to protect endangered species
D) give more money to the poor countries without increasing education
Biology
2 answers:
kirza4 [7]3 years ago
8 0

The answer to this question is A.

lbvjy [14]3 years ago
3 0
D.
...........................
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Polysaccharide

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Scientific investigations begin by asking questions that can be tested, measured, or both. Which question meets this criterion?
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The answer choices are not provided, so I cannot select the correct one. However, I may provide you examples and guidance on how to choose the correct one.
As mentioned in the explanation, a scientific question must have a variable that is changed, the independent variable, and a variable on which the effect of the change is measured. This is the dependent variable. Examples of such questions include:

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During photosynthesis energy from the sun is converted into
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In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa
Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) <em>9/16 D_E_  wild type </em>

<em>     3/16 D_ee  wild wing, ebony body</em>

<em>     3/16 ddE_ dumpy wing, wild body</em>

<em>     1/16 ddee dumpy wing, ebony body</em>

(b) Chi square X^2 = \frac{(O - E)^2}{E}, where O = observed frequency and E = expected frequency.

Phenotype                O                  E                                           X^2

Wild type                 473          9/16 x 830 =  466.875     \frac{(473 - 466.875)^2}{466.875} = 0.08

wild w/ebony b      156          3/16 x 830 = 155.625       \frac{(156 - 155.625)^2}{155.625} =  0.0009

Dumpy w/wild b     149          3/16 x 830 = 155.625      \frac{(149 - 155.625)^2}{155.625} = 0.28

dumpy w/ebony b  52            1/16 x 830 = 51.875        \frac{(52 - 51.875)^2}{51.875} = 0.0003

Total X^2 = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of X^2 (0.05)= 7.815

(c) <em>Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.</em>

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