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Ludmilka [50]
3 years ago
13

Name two challenges that scientists may encounter when they’re monitoring populations of organisms in the wild.

Biology
2 answers:
velikii [3]3 years ago
4 0

One challenge is cost. To effectively monitor populations in the wild, scientists need special equipment such as cameras, computers, and GPS monitoring equipment. Another challenge is accuracy. It’s impossible to count every single organism in an ecosystem, so scientists often work with estimates.


Phoenix [80]3 years ago
3 0

1 - Weather. If scientists are monitoring organisms in their natural environments, than the weather conditions may not be fit to their equipment, clothing, etc. For instant, if it was a rainforest where rain is common, a challenge would be to get good visuals on the organisms.

2 - Aggressiveness. If scientists were to monitor organisms that are new or that they know nothing about, they could unintentionally aggravate the organism or those around them in the wild. In this situation, the aggravation could lead to destruction of data or equipment and lead to a possible set up in the observation.

Hope this helps!

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A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of
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- 100 grams of food A (x)

- 60 grams of food B (y)

Explanation:

According to this question, the group of laboratory rats are being fed a daily diet of:

- 125.6 units of niacin and 5,400 units of retinol

Let x represent the number of grams needed for food A.

Let y represent the number of grams needed for food B.

This means that:

Food A: Niacin = 0.86x, Retinol = 30x

Food B: Niacin = 0.66y, Retinol = 40y

Since there are 125.6 units of niacin fed to the rats in total, then;

0.86x (A) + 0.66y (B) = 125.6

Since there are 5400 units of retinol fed to the rats in total, then;

30x (A) + 40y (B) = 5400

Solving these two equations simultaneously using elimination method:

0.86x (A) + 0.66y (B) = 125.6 ........(eqn 1)

30x (A) + 40y (B) = 5400 ............ (eqn 2)

Multiply eqn 1 by 30 and eqn 2 by 0.86

30 × 0.86x + 0.66y = 125.6

0.86 × 30x + 40y = 5400

25.8x + 19.8y = 3768 ............ (eqn 3)

25.8x + 34.4y = 4644 ........... (eqn 4)

Subtract eqn 3 from eqn 4

14.6y = 876

Divide both sides by 14.6

y = 876/14.6

y = 60

Substitute value for y (60) into eqn 2

30x + 40y = 5400

30x + 40(60) = 5400

30x + 2400 = 5400

30x = 5400 - 2400

30x = 3000

x = 3000/30

x = 100

Therefore, she feeds the group of rats 100grams of food A (x) and 60 grams of food B (y).

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