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shutvik [7]
3 years ago
8

What is a difference between prokaryotic and eukaryotic cells?

Biology
1 answer:
Komok [63]3 years ago
7 0

Answer:

b

Explanation:

b beaacs it sound better

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Dont mind the answer I picked
Marat540 [252]

Answer:

cold water temperature of the lake

Explanation:

mark me branilist

4 0
2 years ago
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What is the 95% CI for data set 2?
vesna_86 [32]

Answer: 20 ± 11.32

Explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values where there is a % certainty the true mean value lies in. For example, a 95% CI means there is a 95% certainty the true mean value lies in the interval.

For the data set 2, z-value for 95% confidence interval is z = 1.96

Formula to calculate CI is

x ± z\frac{s}{\sqrt{n} }

where

x is mean

z is z-value

s is standard deviation

n is the sample number

Then:

20 ± 1.96\frac{10}{\sqrt{3} }

20 ± 1.96\frac{10}{1.7320}

20 ± \frac{19.6}{1.7320}

20 ± 11.32

The 95% CI is 20 ± 11.32, which means there is a 95% certainty the true value is between 8.68 and 31.32.

8 0
3 years ago
What is a solstice?
nevsk [136]

Answer:

4 or D

Hope it’s right ;)

5 0
2 years ago
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In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the freq
DENIUS [597]

Answer:

If the frequency of the dominant allele in the pea population is 0.8, the genotype frequencies in the population would be 0.64 AA, 0.32 Aa, and 0.04 aa.

Explanation:

Hardy-Weinberg equation tells us that:

p2 + 2pq + q2

p2 is the frequency of AA plants

2pq is the frequency of Aa plants, and

q2 the frequency of aa individuals.

A allele in the population is 80%, the frequency is 0.8.

p represent the frequency of A allele. p is 0.8

Therefore calculating for q ( frequency of a allele).

p + q = 1.0

q = 1.0 − p

q = 1.0 − 0.8

q = 0.2

p = 0.8, q = 0.2.

Now we can calculate the predicted frequencies of the different genotypes, remembering that p2 is the frequency of the AA genotype, 2pq is the frequency of the Aa genotype, and q2 the frequency of aa genotype.

p2 = 0.8 x 0.8 = 0.64

2pq = 2 x 0.8 x 0.2 = 0.32

q2 = 0.2 x 0.2 = 0.04.

Thus, we would expect to see genotype frequencies of 0.64 AA, 0.32 Aa, and 0.04 aa in the pea plants.

6 0
3 years ago
What type of bonds does carbon atoms form?
Tamiku [17]

Answer:

carbon bond is a covalent bond between two carbon atoms.

Explanation:

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