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adelina 88 [10]
3 years ago
15

Which greenhouse gas(es) in the table have carbon in them? Choose all that apply

Biology
1 answer:
stealth61 [152]3 years ago
3 0
A) Carbon Dioxide just took the test
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Which change to the ramp would be the best way to increase its medical efficiency (ME), and why?
olchik [2.2K]

Answer:

make it lower

Explanation:

because then it would be easier to maneuver

3 0
3 years ago
Which of these gymnosperms are known to be important in maintaining the balance of carbon in an ecosystem?
CaHeK987 [17]

The correct answer is:     [D]:

___________________________________________________

                    →   " redwood tree forests " .

___________________________________________________

 <u>Note</u>:  Let us consider the other answer choices.

___________________________________________________

Choice:  [A]:  "garden flowering plants" ;  is incorrect  — since these are not generally considered to be part of a true or traditional "ecosystem" ; and especially since "garden flowering plants" are, in fact, "flowering plants" —  which means that they are "angiosperms" —  <em><u>Not</u></em>  "<em><u>gymnosperms</u></em><em> </em>".

      →  Rule out:  " Choice:  [A] " .

__________________________________________________

Choice:  [B]:  " ferns and liverworts " ; is incorrect — since <em><u>both</u></em>  ferns — as well as liverworts —<u><em> are Not  </em></u><u>"</u><u><em>gymnosperms </em></u>".  

          Liverworts are "bryophytes" {which lack true vascular tissue] ; and ferns are "pterophytes" ["seedless vascular plants"].

      →  Rule out:  " Choice:  [B] " .

___________________________________________________

Choice:  [C]: " mosses and horsetails "; is incorrect ;  since <em><u>both</u></em> mosses — as well as horsetails — <u><em>are Not  </em></u><u>"</u><u><em>gymnosperms </em></u>".   Mosses are "bryophytes" (see aforementioned) ; and horsetails are "non-seed bearing plants" that spread vegetatively [underground; by asexual reproduction — <em><u>and Not by seed</u></em><em> .</em>].

      →  Rule out:  " Choice:  [C] " .

__________________________________________________

Choice:  [D]: " redwood tree forests "  ;

               — is the only answer choice provided that is a "gymnosperm".    

             Furthermore, the "redwood tree" is a conifer (i.e. "coniferous; or, "cone-bearing" plant} that reproduces through seeds spread by "cone") — and all "conifers" are a type of "gymnosperm."  

            Furthermore, "redwood trees" grow large and tall, and redwood forest communities within an ecosystem; due to there size and presence, are:  "<em><u>known to be important in maintaining the balance of carbon in an ecosystem</u></em>."

__________________________________________________

 →  As such, the correct answer is:

__________________________________________________

 →  Answer choice:  [D]:  " redwood tree forests " .

__________________________________________________

Hope this helps!

    Best wishes to you in your academic endeavors

           — and within the "Brainly" community!

__________________________________________________

5 0
3 years ago
Read 2 more answers
In Figure 12–5, what nucleotide is going to be added at point 2, opposite from guanine?
Luden [163]
Cytosine is the answer
3 0
3 years ago
Read 2 more answers
A population is made up of individuals where 77 have the A1A1 genotype, 65 have the A1A2 genotype, and 123 have the A2A2 genotyp
lana [24]

Answer:

Frequency of allele A1- 0.41

Explanation:

In Hardy weinberg equilibrium,

P refers to the dominant allele

q refers to the recessive allele

The allele frequency will be p+q=1

The genotypic frequency is- P²+q²+2pq=1

P²= genotype of dominant trait ( A1A1)- 77

2pq= genotype of heterozygotes (2pq)- 65

q²= genotype of recessive trait (A2A2)-  123

Total number of offsprings= 77+  65+ 123

                                            = 265

Now to calculate allele frequency of A1=

\dfrac{\text{number of A1A1 offspring}}{ \text{total number of offspring}}+\dfrac{1}{2}(\dfrac{\text{number of A1A2 offspring}}{\text{total number of offspring}})

= 77/265 + 1/2( 65//265)

=  0.290+ 0.122

= 0.413

Thus, 0.41 is correct.

3 0
3 years ago
What class of cells are bacteria apart of
lana66690 [7]
They are prokyartic cells 
8 0
3 years ago
Read 2 more answers
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