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saw5 [17]
3 years ago
15

Give three examples of how this tree could affect

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

Answer:If a forest fire in Florida cleared thousands of acres of trees the consequences for the animal populations in the surrounding area would be animal populations will increase in the surrounding forests.

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Which of the following occurs in a right lateral movement? A. The right condyle primarily rotates. B. The right condyle moves do
Thepotemich [5.8K]

Answer:

"The right condyle primarily rotates" occurs in a right lateral movement

Explanation:

The mandibular condyle is dynamically participate in endochondral ossification. The condylar cartilage is an significant development place in the mandible, paying to the elongation of the mandibular ramus. A protrusive energy happens when the mandible deviates (both of condyles) onward and descending along the slope of the articular eminence Happens though cutting and grasping food. This shift happens subsequently the condyle rotates. Regular motion of the mandible relies on appropriate role of the TMJ. Visibly, the preauricular area lies straight over the joint.

3 0
4 years ago
Explain why osmosis is important in plant and animal​
olga2289 [7]

Answer:

Osmosis is important in plants and animals because it allows for the absorption of water.

Explanation:

7 0
4 years ago
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The first electric generators used direct current, so they needed to be reversed manually on a regular basis in order to work pr
Rashid [163]
<span>B.Tesla used another scientist’s research as inspiration.</span>
5 0
3 years ago
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In a one-group pretest-posttest design affected by the problem of testing effects, the reduction found on the posttest is likely
Kisachek [45]

Answer:

taking the pretest

Explanation:

A one-group pretest–posttest design is a type of research design generally used in behavioral research aimed at estimating the treatment effects on the sample. In this experimental design, the target group of participants is first pre-tested on the dependent variable and, subsequently, post-tested after the administration of a treatment. In consequence, the participants of the target group receive the same treatment and assessment. The problem of testing effects indicates that testing may become a problem when the pretest may change participant's behavior. ​When there are statistically significant differences between the pretest and posttest data, it is possible to support the working hypothesis.

7 0
3 years ago
22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

X^2  =  \frac{(O - E)^2}{E} where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The X^2 is calculated thus:

Phenotype            O               E                             X^2

  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

4 0
3 years ago
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