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AlexFokin [52]
3 years ago
8

How does the type of rock in Earth's crust affect the distribution of groundwater

Biology
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

Obviously, such rocks transmit only small quantities of water and are poor aquifers. By comparison, rocks such as fractured sandstones and cavernous limestone have large connected openings that permit water to move more freely; such rocks transmit larger quantities of water and are good aquifers.

Explanation:

Olegator [25]3 years ago
4 0

Answer:

Rocks transmit small quantities of water and are poor aquifers. Comparison, rocks such as fractured sandstones and cavernous limestone have large connected openings that permit water to move more freely.

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Paddle like swimmerets are used for movement by crustaceans.

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Which statement describes potassium-argon dating?
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Answer: Potassium-40 decays into argon gas over time.

Explanation: Potassium-argon dating is a dating method used to determine the age of sedimentary rocks by comparing the proportion of K-40 to Ar-40 in a sample of rock, and knowing the decay rate of K-40.

Potassium-40 undergoes decay following first order kinetics as given below:

_{19}^{40}\textrm{K}\rightarrow _{18}^{40}\textrm{Ar}+_{0}^1e

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The vestibular system is the only sensory system to send projection fibers directly to the cerebellum.
Greeley [361]

Answer: False

Explanation:

The above statement is false as, cerebellum receives the projection fibers from the other sensory system in which they are received in the different part of cerebellum. As, the cerebellum received the information from the sensory system and the spinal cords and also in the other part of the brain as it regulates in the movement of the motor.

3 0
3 years ago
The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A
ratelena [41]

Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

<em><u>Genotype       Observed frequency</u></em>

    A             →        18 individuals

    B             →        55 individuals

    C             →        27 individuals

Do these contradict the genetic model?  

Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.  

Explanation:

<u>Available data</u>:

  • Crossed genotypes: two
  • Genotypes among the offspring; Three → A, B, and C
  • Expected phenotypic ratio → 1:2:1
  • Total number of individuals, N = 100
  • A = 18 individuals
  • B = 55 individuals
  • C = 27 individuals

So, let us first state the hypothesis:

  • H₀= the population is equilibrium for this locus → F(A) = 25%,  F(B) = 50%, F(C) = 25%  
  • H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 ---------------  25% -------------X = 25 individuals A

2 --------------  50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

<u>                                                   A                             B                           C</u>

  • Observed                         18                            55                         27
  • Expected                         25                           50                         25
  • (Obs-Exp)²/Exp                1.96                        0.5                        0.16

<u>(Obs-Exp)²/Exp</u>

A)  (18 - 25)²/25 = 49/25 = 1.96

B)  (55 - 50)² / 50 = 25/50 = 0.5

C)  (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp  

  • ∑ is the sum of the terms
  • O are the Observed individuals: 2 in chamber B, and 18 in chamber A.  
  • E are the Expected individuals: 10 in each chamber  

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05  

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99    

<em>These results suggest that there is </em><u><em>not enough evidence to reject</em></u><em> the null hypothesis. We can assume that </em><u><em>the locus under study in this population is in equilibrium H-W.  </em></u>

 

6 0
3 years ago
Which animal adaptation would be best for this habitat?
ASHA 777 [7]

Answer:

it's D. I believe anyways is it if you can pick two pick A,D, or D and C who knows at this point

sorry I'm not much help but I hope it did something

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3 years ago
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