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Fantom [35]
3 years ago
15

Why does a chameleon have a broad niche? explain PLEAEEEEEE NEED HELP ASAP!!!!!!!!

Biology
1 answer:
mart [117]3 years ago
6 0
That’s the answer hope it helps!!!

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Help please thanks This is for Science
Drupady [299]

Answer:

no he got banned

Ex

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5 0
3 years ago
Which of the following is not a freshwater ecosystem?
9966 [12]

Answer:

c

Explanation:

seas water isn't fresh water

4 0
3 years ago
Read 2 more answers
El cloroplasto es el conducto de la fotosíntesis<br>​
jolli1 [7]

Answer:

Bonjour

Explanation:

salut je ne connais pas cette langue désolé

8 0
3 years ago
In a population of 150 toads, poisonous (P) is dominant over non-poisonous (p). According to the Hardy-Weinberg equation, the ex
aalyn [17]

Answer: Χ²=8.18

Explanation: In this population, the frequency of homozygote poisonous toads is:

p^{2} =\frac{100}{150}

p^{2} = 0.667

p = \sqrt{0.667}

p=0.82

For the homozygote non-poisonous toads:

q^{2} = \frac{5}{150}

q^{2} =0.034

q = \sqrt{0.034}

q = 0.18

The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: p^{2} + 2pq + q^{2} = 1, in which, heterozygote frequency is given by 2pq

2pq = 2*0.82*0.18 = 0.3

Now, to compute the chi-square test, follow the instructions:

1) Find the observed values: in this case, they are the found frequency:

0.82       0.3         0.18

2) Find the expected values: As the question mentioned, the expected proportion is:

16            8             1

3) Subtract the observed value from the expected value:

0.82 - 16 = - 0.184          0.3 - 8 = -7.7             0.18 - 1 = - 0.818

4) Square each value from above:

(-0.184)² = 0.034            (-7.7)² = 59.3                 (-0.818)² = 0.77

5) Divide each value by expected value:

\frac{0.034}{16} = 0.0021                   \frac{59.3}{8} = 7.41                      \frac{0.77}{1} = 0.77

6) Add all the values and we will have the chi-square test:

Χ² = 0.0021 + 7.41 + 0.77 = 8.18

The chi-square test is Χ² = 8.18

8 0
3 years ago
Which of the following results provided evidence of a discrete nuclear localization signal somewhere on the nucleoplasmin protei
Alja [10]

After cleavage of the nucleoplasmin protein, only the tail segments appeared in the nucleus is evidence of a discrete nuclear localization signal somewhere on the nucleoplasmin protein.

7 0
4 years ago
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