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adoni [48]
3 years ago
7

Is aristotle a real scientist

Biology
1 answer:
monitta3 years ago
8 0
Yes, he is a philosopher and scientist. Born 384 BC died 322 BC.
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Worth 26 points! Help me please! Do 1,2,3 by filling in the blank!!!! And I will make you brainest!
sergij07 [2.7K]

Answer:

Not to sure but try this...

1st blank try (white dots)

2nd blank try (black dots)

3rd blank try (red dots)

Hope this helps!!!

Have a good night!!!! :)

6 0
3 years ago
Read 2 more answers
Why are forests called "biodiversity hot spot"?
IrinaK [193]

Answer:

Explanation: It is a region of the earth that is extremely biological diverse and also

Under many threat due to habitat loss , clamate change or extensive species for guarding these region is important because host animals are there which is not found anywhere on earth

8 0
3 years ago
PKU (Phenylketonuria) is an autosomal recessive disease, in which the synthesis of amino acid Tyrosine from Phenylalanine is blo
prisoha [69]

To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.

If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.

p +q=1

0,3 +q=1

q= 1-0,3

q= 0,7

Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.

p^{2} + 2pq + q^{2} = 1

(0,3)^{2} + 2pq +(0.7)^{2} = 1

0,09 + 2pq +0.49= 1

2pq +0,58= 1

2pq= 1-0.58

2pq=0,42

The freqency of the heterozygotes in this population is 0.42

6 0
3 years ago
Devise an experiment to test the hypothesis that bright orange coloration in butterflies acts as a "warning" to potential
Murrr4er [49]

Answer and Explanation:

As I don't have access to live butterflies, the control group would be clay models that do not exhibit warning coloration, e.g. beige / white / brown. In contrast, the experimental group would be the clay models that exhibir a strong warning coloration such as color orange / red.

After preparing the clay 'prey', I would locate them in specific places outside where birds are able to detect them. Moreover, I would use a webcam to monitor the recurrence of birds in the area.

1st experiment: Test whether birds approach the prey that exhibits both warning coloration (experimental group) and dull coloration (control group). Then, record the minutes they spent wandering near the prey.

2nd experiment: Test wheter birds feed on the prey that exhibits both warning coloration (experimental group) and dull coloration (control group). Record time.

Carry out statistical analyses.

Present results and discuss.

3 0
3 years ago
Science: Match the parts of the eye to their functions: Structure: 1. Iris 2. Cornea 3. Lens 4. Retina 5. Photoreceptors 6. Opti
alexgriva [62]

Answer:

iris - b

cornea - f

lens - e

retina - c

photoceptors - b

optic nerve - h ( definately know dis one)

rod cells - e

pupil - a

cone cells -g

i tried

Explanation:

5 0
3 years ago
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