Inflammation of a vein is called<u> phlebitis</u>.
In the field of biology, veins can be described as a type of blood vessel whose major function is to collect deoxygenated blood from all the parts of the body and travel it to the heart.
An inflammation of the vein, referred to as phlebitis, can be a painful experience and there are a number of reasons for a vein to become inflamed.
Phlebitis might occur because the blood in a vein has become clogged due to a certain reason. It might also happen that the walls of the vein become damaged due to which phlebitis occurs.
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Tertiary Consumers, then decomposers, once the tertiary consumers die
Answer:
I'm going to have to say 2
Explanation:
I'm not sure about photosynthesis because it's not making any glucose, or it doesnt show that in the diagram, but it is making oxygen which is a chemical reaction because it cant go back to carbon dioxide through the plant.
Complete question:
The Hardy-Weinberg equilibrium is a theoretical model that predicts the frequency of different genotypes in a population that is not evolving. A scientist studying the population genetics of the rainbow fish finds that the Hardy-Weinberg model predicts that the frequencies of the three different genotypes for the albino gene in a given lake should be as follow: AA: 49%, Aa: 42%, and aa: 9%. However, the scientist obtained a sample of 320 individuals of this species and found that the numbers of individuals for the different genotypes are: AA: 180, Aa: 120, and aa: 20. Use the appropriate statistical technique to test (use alpha = 0.05) whether the observed genotype frequencies are consistent with eh Hardy-Weinberg model. State your conclusion.
Answer:
There is enough evidence to reject the null hypothesis. The genotypes are not in equilibrium. The genotype frequencies are consistent with the Hardy-Weinberg model.
Explanation:
Expected Number of individuals with each genotype: Each genotype´s frequency multiplied by the total number of individuals in the sample.
AA= 0.49 x 320 = 156.8
Aa= 0.42 x 320 = 134.4
aa= 0.09 x 320 = 28.8
Observed Number of individuals with each genotype:
AA= 180
Aa= 120
aa= 20
Chi square= ∑ ((O-E)²/E)
- ∑ is the sum of the terms
- O are the Observed individuals
- E are the Expected individuals
AA= (O-E)² /E
AA= (156.8 - 180) ² / 156.8
AA= 538.24/156.8
AA= 3.433
Aa= (O-E)² /E
Aa= (134.4-120)² / 134.4
Aa=207.36/134.4
Aa= 1.543
aa= (O-E)² /E
aa= (28.8-20)²/28.8
aa= 2.69
X² = ∑ ((O-E)²/E) = 3.433 + 1.543 + 2.69 = 7.666
Freedom degrees = genotypes - alleles = 3 - 2 = 1
Significance level, 5% = 0.05
p value less than 0.05
Table value/Critical value = 3.841
7.666 > 3.84 meaning that the difference between the observed individuals and the expected individuals in each chamber is statistically significant.
There is enough evidence to reject the null hypothesis. The genotypes are not in equilibrium.
Answer: External intercoastal muscles increase the size of thorasic cavity during inhalation.
Explanation:
Inhalation is the process of breathing in oxygen.
During inhalation, diaphragm contracts and it lead to the expansion of lung volume. When diaphragm contracts, it moves to the abdominal cavity which create a large thorasic cavity . when the external intercoastal muscles contracts, it cause toward and downward movement of the ribs, which cause the rib cage to expand , thereby leading to increase in size of thorasic cavity.
