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Evgesh-ka [11]
3 years ago
12

What's an animal that eats large amounts of roughage but does not have a stomach with several compartments?

Biology
1 answer:
Greeley [361]3 years ago
6 0

Answer:

ruminants

Explanation:

A pseudo-ruminant is an animal that eats large amounts of roughage but does not have a stomach with several compartments. The digestive system does some of the same functions as those of ruminants.

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<span><span>Lipids are the class of organic compounds that stores energy as fat</span>.

The answer is Lipids.

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Hi can ya'll pls do my homework lol I'm too lazy. I'll give you brainliest anddd be rlly happy?
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Answer:

Bet

Explanation:

1. The radiaton from the X-ray waves can be harmful

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3 years ago
These symptoms are important in fighting off infection ?
VARVARA [1.3K]

It’s A the immune system & lymphatic Hope this helps!

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Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
Sphinxa [80]

Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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Description of Vegetative​
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Answer:

Vegetation def is: growing plants, or a life without physical, mental or social activity

Vegetative depends on the context. Mainly just means without brain activity

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