Answer:
Taking more samples from different parts of an acre.
Explanation:
Validity and accuracy of the data is crucial for any serious research. In this particular case, the most accurate data would be obtained if earthworms would be counted on the whole acre. Of course, this would consume lot of time, people, money etc. That's why methods for estimation are used. Estimation best works with large number of samples. Since one acre equals over 4000 square meters, taking only five samples from such a big area is simply not enough for obtaining valid data.
One of the possible ways to improve estimation is to take more samples per acre while avoiding taking adjacent samples because it could be possible that number of earthworms in one part of an acre is increased (or decreased) for any reason, which would lead us to wrong conclusion.
<span>1. Pose significant questions that can be investigated empirically.
2. Link research to relevant theory.
3. Use methods that permit direct investigation of the question.
4. The methods used to obtain data and test hypotheses should be benevolent and not malevolent.</span>
Answer:
4. A
4.(part 2) B
6. C
7. A
8. C
9. A
10. B
11. C
12. D
13. C
14. D
15. C
16. C
17. A
18. C
19. B
20. G
I Really hope this was helpful to you i tried my best to answer all the questions
Explanation:
Answer:
The rabbit that best adapted to its environment was the gray rabbit.
Explanation:
Answer:
1/8
Explanation:
Given that the trihybrid parents have AaBbCc genotype for fruit color. The trait is a quantitative trait i.e. each dominant allele will have an additive effect on it. In this case, AaBbCc and AABBCC will not produce same fruit color because AaBbCc has only three loci contributing to the color while in AABBCC all the six loci are contributing to the color. For an offspring to be exactly similar to the AaBbCc parents it should have the same genotype of AaBbCc.
The probability of Aa to come from a cross between Aa and Aa is 2/4 or 1/2
The probability of Bb to come from a cross between Bb and Bb is 2/4 or 1/2
The probability of Cc to come from a cross between Cc and Cc is 2/4 or 1/2
So the collective probability of AaBbCc offspring from a cross between AaBbCc and AaBbCc parents would be=
1/2 * 1/2 * 1/2 = 1/8
Hence, assuming no effects of the environment, 1/8 of the offspring will have the same fruit color phenotype as the trihybrid parent.
