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siniylev [52]
3 years ago
6

Why do you want to read research by scientists who use the scientific method? Select all that apply.

Biology
1 answer:
Bezzdna [24]3 years ago
8 0

Answer: I believe it is C. and D.

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Explain living things such as people in trees and how they are different from nonliving things such as rocks and tents
Citrus2011 [14]

Answer:

The difference between living things and nonliving things is little things that we call is senses like sight, taste, touch, smell, and hearing. For example a human or animal can react to someone or something touching them. For a nonliving thing such as a rock can't react to someone touching them unless u force them to.

Explanation:

https://quizlet.com/120523212/characteristics-of-life-biology-identify-the-levels-of-organization-in-a-multicellular-organism-flash-cards/

6 0
3 years ago
26. Analyze In a particular species of butterfly, the genes
Reil [10]

Answer:

law of independent assortment

Explanation:

Independent assortment is the random distribution of the pairs of genes on different chromosomes to the gametes.  I hope this helps :)

4 0
3 years ago
The small flap of cartilage that folds over the windpipe when you swallow.
Rainbow [258]
<span>uvula is the answer for this problem.</span>
8 0
3 years ago
Select the correct text in the passage.
Helga [31]

Answer:

A construction company has been called in to divert all excess water flowing into the pond to an unused natural reserve 800 feet

away.

Explanation:

Everything else is smoke and mirrors or just irrelevant

7 0
3 years ago
Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this
ad-work [718]

The genealogy tree which is missing from the question is in the attachment.

Answer: 1) P1 = \frac{1}{2}; 2) P2 = \frac{1}{2}; 3)P3 = \frac{1}{16}

Explanation: 1) According to the genealogy tree, the individual III-3 has a P1 = 1/2, because each of the parents are heterozygous for the condition, since one of the ofspring is affected.

2) For individual III-4, one of the parents is heterozygous for the condition, so the probability of being a carrier is P2 = 1/2.

3) Now, for individual IV-1 be affected, the parents has to be carrier. So, this probability, depends on the probability of each parent being heterozygous and that the individual will be carrying the condition, which means:

P3 = 1/2 · 1/2 · 1/4 = 1/16.

The probability of having the condition is 1/16 or 6.25%.

4 0
2 years ago
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