Answer:
Because of the mountaineous terrain and rocky landscape
Explanation:
Pls vote my answer as brainliest
Answer:
B. are more common in natural systems altered by human actions
Explanation:
Positive feedback loops: The term "positive feedback loops" is referred to as a phenomenon that is responsible for amplifying or enhancing changes and therefore due to this, a system moves away from the state of equilibrium state to make it unstable on a large scale.
In the question above, positive feedback loops are considered more common in the natural systems that are being altered by human actions.
Example:
1. When an employee models good behavior.
2. When a colleague meets or exceeds goals.
Answer:
Respuesta: Debemos eligir de forma responsable nuestro gobierno escolar porque serán las autoridades que verán por los intereses de la institución educativa y sus estudiantes, por lo que deben ser fieles a los valores y principios que profese dicha institución.
Explanation:
When centrifuging biohazards, it is recommended that the safety buckets and sealed rotors be: Loaded and unloaded within the biosafety cabinet.
<h3>What is meant by centrifuging biohazards?</h3>
This is the term that has to do with the user of centrifugal force in the separation and the removal of substances in liquid as well as in solid materials.
What is required is that When centrifuging biohazards, it is recommended that the safety buckets and sealed rotors be: Loaded and unloaded within the biosafety cabinet.
Read more on Biohazards here: brainly.com/question/10686710
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Answer:
Explanation:
Your question has one part only: <em>a) The average weight of the eggs produced by the young hens is 50.1 grams, and only 25% of their eggs exceed the desired minimum weight. If a Normal model is appropriate, what would the standard deviation of the egg weights be?</em>
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<h2><em>Solution</em></h2><h2><em /></h2>
You are given the <em>mean</em>, the reference value, and the <em>percent of egss that exceeds that minimum</em>.
In terms of the parameters of a normal distribution that is:
- <em>mean</em> =<em> 50.1g</em> (μ)
- Area of the graph above X = 51 g = <em>25%</em>
Using a standard<em> normal distribution</em> table, you can find the Z-score for which the area under the curve is greater than 25%, i.e. 0.25
The tables with two decimals for the Z-score show probability 0.2514 for Z-score of 0.67 and probabilidad 0.2483 for Z-score = 0.68.
Thus, you must interpolate. Since, (0.2514 + 0.2483)/2 ≈ 0.25, your Z-score is in the middle.
That is, Z-score = (0.67 + 0.68)/2 = 0.675.
Now use the formula for Z-score and solve for the <em>standard deviation</em> (σ):
