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Gemiola [76]
3 years ago
11

The body of arboreal animals is flexible.why​

Biology
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

the body of arboreal animals is flexible that enables them to pass through narrow branches.

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Which stage of cellular respiration produces 32 or 34 atp
Leni [432]
Krebs cycle is the stage that produces 32 or 34 atp
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3 years ago
Both animals and fungi are heterotrophic. What distinguishes animal heterotrophy from fungal heterotrophy is that only animals d
DiKsa [7]

Answer:

Heterotrophs are the organisms which cannot prepare their food on their own like autotrophs instead are dependent on other mode to obtain the organic substances from the environment. This method is common in animal and fungi groups.

Although both fungi and animals are heterotrophs their mode of heterotrophic is different as fungi obtain organic substance b secreting many digestive enzymes which digests the complex biomolecules and the fungi obtain the organic substances whereas the animals swallow or ingest the material and then digests it.

6 0
3 years ago
How can a robin have a ddt concentration of 444 ppm even though the soil contains only 10 ppm of ddt
zaharov [31]

Robin can have a ddt concentration of 444ppm while the ddt concentration of soil is 10 ppm. This happens due to biological magnification.

<u>Explanation:</u>

In biological magnification the concentration of toxic substances at successive trophic levels keeps increasing. Due to biomagnification, what starts out as a small concentration grows  while moving to successive trophic levels. DDT is a pesticide which is used to kill pests that affect crops.

DDT used in farms gets deposited in the soil. From the soil it can reach various trophic levels. At each trophic level its concentration keeps increasing.

Robin is a bird that is a secondary consumer and thus belongs to the higher trophic level. Thus even if the DDT concentration is as low as 10ppm in the soil its concentration will keep increasing and can become a high value like 444ppm.

3 0
3 years ago
Select ALL the correct answers.
tekilochka [14]

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}

10th\ run = 9

So:

B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}

Required

Select all true statements

(a) & (d) Median Comparisons

A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}                         B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}

n = 9                                                         n = 10

Arrange the data:

A = \{1, 1, 1, 1, 2, 2,  2,3, 4\}               B = \{1,1,1,1,2,2,2,3,4,9\}

                               Median = \frac{n + 1}{2}th

Median = \frac{9 + 1}{2}th                            Median = \frac{10 + 1}{2}th

Median = \frac{10}{2}th                              Median = \frac{11}{2}th

Median = 5th                                 Median = 5.5}th --- average of 5th and 6th

Median = 2                                    Median = \frac{2+2}{2} = 2

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

A = \{1, 1, 1, 1, 2, 2,  2,3, 4\}               B = \{1,1,1,1,2,2,2,3,4,9\}

n = 9                                                         n = 10

First, calculate the lower quartile (Q1)

Q_1 = \frac{n + 1}{4}th[Odd n]             Q_1 = \frac{n}{4}th [Even n]

Q_1 = \frac{9 + 1}{4}th                            Q_1 = \frac{10}{4}th

Q_1 = \frac{10}{4}th                              Q_1 = 2.5

Q_1 = 2.5th                              

This means that:

Q_1 = 2nd + 0.5(3rd - 2nd)              Q_1 = 2nd + 0.5(3rd - 2nd)

Q_1 = 1 + 0.5(1- 1)                   Q_1 = 1+ 0.5(1 - 1)                      

Q_1 = 1                                       Q_1 = 1

Next, calculate the upper quartile (Q3)

Q_3 = \frac{3}{4}(n + 1)th [Odd n]             Q_3 = \frac{3}{4}(n)th [Even n]

Q_3 = \frac{3}{4}(9 + 1)th                            Q_3 = \frac{30}{4}th

Q_3 = \frac{30}{4}th                                     Q_3 = 7.5th  

Q_3 = 7.5th                                    

This means that:

Q_3 = 7th + 0.5(8th- 7th)           Q_3 = 7th + 0.5(8th- 7th)

Q_3 = 2 + 0.5(3- 2)                       Q_3 = 2+ 0.5(4 - 2)                      

Q_3 = 2.5                                       Q_3 = 3

The interquartile range is  IQR = Q_3 - Q_1

So, we have:

IQR = 2.5 - 1                  IQR = 3 - 1

IQR = 1.5                       IQR  =2

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

8 0
3 years ago
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(4) amount of protein digested

8 0
3 years ago
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