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Lesechka [4]
3 years ago
12

Sarah is documenting her sources. Which step of research process is she on? A. Step 5: Cite Your Sources B. Step 4: Organize You

r Information C. Step 2: Find and Retrieve Materials D. Step 3: Evaluate Your Sources
Biology
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

Either A or B

Explanation:

kiruha [24]3 years ago
6 0

Answer: A. Step 5: Cite Your Sources

Explanation:

A citation is a documentation and recognition of the research work and references used for the purpose of present research. Apart from research work other original sources such as idea, images and information can be used for the purpose of reference. In the body of a paper,small acknowledgments of the source of information used is required to be mentioned. In the end of the research article or thesis the entire list is framed for the references of the authors or other researchers.  

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MAVERICK [17]

Answer:

The correct answer is option (d) "both b and c".

Explanation:

Experimental model organisms used in scientific investigations have to fulfill certain criteria in order to be useful in the laboratory. Two of the desired characteristics are being able to reproduce quickly and with enough number of offspring in order to study genetic behaviors. <em>Arabidopsis thaliana</em> characteristics of being able to reproduce in 8–10 weeks and producing thousands of offspring per plant (because it can produce thousands of seeds) makes it a model flowering plant for experimental studies.

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Answer:

A) Humus

Explanation:

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Suggest why the lynx populations appear to be so dependent on the fluctuations on the hare
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Which best describes the difference between osmosis and diffusion?
joja [24]
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~Lauv
6 0
4 years ago
Read 2 more answers
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
ArbitrLikvidat [17]

Answer and Explanation:

a) Frequencies of A and G lalleles are as follows:

f(A) =( 84+24)/174 = 0.62

f(B) = (42+24)/174 =0.38

b) Expected genotype frequencies:

f(AA) = (0.62) (0.62) = 0.384

f (AG)= 2(0.62) (0.38) =0.471

f(GG) = (0.38) (0.38) = 0.144

c) Genotype Observed Expected O-E (O-E)2 (O-E)2/E

AA 42 33 9 81 2.45

AG 24 41 17 289 7.05

GG 21 13 8 64 4.92

Chi squared = 14.42

The number of degrees of freedom is the number of genotypes minusthe number of alleles= 3-2 =1

The p value is much less than 0.05, therefore we reject the hypothesis that these genotype frequencies may be expected from HArdy Weinberg equilibrum

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3 years ago
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