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Brilliant_brown [7]
2 years ago
6

WILL MARK BRAINLIEST AND GIVE 25 POINTS Explain how you linked the predator and its prey to a specific environment.

Biology
1 answer:
qwelly [4]2 years ago
7 0

Answer:

The mouse and shrew bones were common to all three owls so they were not useful in identifying environments. The gopher bones were unique to the great grey owl of Alaska. Gophers are rodents that are successful in areas with soft soil and lots of grass, so the gopher and this owl would be found in the meadows and evergreen forests of Alaska. The bat was unique to the barking owl of Australia. The bat would be found in woodland habitats where there are many trees the bat can hang from. The gull bones were unique to the short-eared owl of Cuba. Gulls are wetland birds, so the gull and this owl would be found in marshes and coastal areas.  

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What happens to muscles when you stretch them eli5?
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They loosen up to be tighten again
8 0
3 years ago
Which of the following has a true coelom? in. flatworms b. hookworms c. tapeworms d. earthworms
vlabodo [156]

I believe the answer is D. Earthworms.

5 0
3 years ago
An organic will always contain
Naddik [55]

Answer:

Explanation:Compounds which have carbon and hydrogen atoms in their structure or formula are known as organic compounds. For example, a molecule of is organic because it has carbon and hydrogen atom in its formula. ... Thus, we can conclude that an organic molecule will always contain carbon and hydrogen

5 0
3 years ago
You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo
irakobra [83]

Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

4 0
3 years ago
Which term describes cycles of hot and cold temperatures which expand and contract rock in combination with constant wind and wa
marishachu [46]

The correct answer is option C, that is, physical weathering.  

The weathering resulting due to the influences of changing temperatures on a rock, making the rock to split apart is known as physical weathering. There are two main kinds of physical weathering, that is, exfoliation and freeze-thaw.  

Physical weathering takes place mainly in the locations, where few plants grow and little formation of soil occurs, like hot deserts and mountain regions. It takes place either via repeated freezing and melting of water or via contraction and expansion of the rocks surface layer, which is baked by the sun.  


4 0
3 years ago
Read 2 more answers
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