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nirvana33 [79]
3 years ago
5

what is the purpose of a control sample that is known to be from bighead carp in edna surveillance? a. to produce multiple bandi

ng patterns b. to amplify the dna in small samples c. to see if samples contain the target dna d. to identify each species of carp found
Physics
2 answers:
garri49 [273]3 years ago
8 0

Correct answer choice is :


C) To see if samples contain the target dna


Explanation:


In many North American rivers, populations of various kinds of non-native cyprinid fishes are present, including black carp, grass carp, bighead carp, silver carp, common carp, and goldfish. All six of these classes are found in the Mississippi River basin and following their aggression has shown complex, especially where wealth is low. The bighead carp is a species of freshwater fish, one of several Asian carps. It is one of the most intensively exploited fishes in aquaculture, with a yearly global product of over three million tonnes in 2013, mostly from China.

professor190 [17]3 years ago
4 0
I think the correct answer from the choices listed above is option C. The purpose of a control sample that is known to be from bighead carp in edna surveillance is to see if samples contain the target dna. A control sample is a set of sample that <span>does not receive treatment by the researchers.</span>
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<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

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\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

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\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

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\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

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The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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