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Alina [70]
3 years ago
11

When a country awards import licenses on a first-come, first-served basis or on a negotiated basis, taking into consideration a

demonstrated need or worthiness, the country has a:?
Social Studies
2 answers:
ki77a [65]3 years ago
7 0
<span>Answer is Resource-Using procedures - first-come, first-serve; basis of demonstrating need or worthiness; basis of negotiation. some of area c is lost to the country</span>
Sonbull [250]3 years ago
4 0
The answer to this question is <span>resource-using procedure.
Import liscense on first-come first-serve basis will defintiely limit the overall total import toward that country.
One reason why a country use this procedure is to make sure that the nation shall efficiently use all of available resources in that country before relying on outside sources to fulfill the demand of the market.</span>
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Name the states that were later carved out of land brought lnto the unión by the louisiana purchase
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Carolyn's therapist told her to relax and spontaneously say whatever thoughts or images came to her mind. Carolyn's therapist wa
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3 years ago
Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller​ eggs, often weighing less than
miv72 [106K]

Answer:

          \large\boxed{\large\boxed{1.33g}}

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>

<em />

<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)
  • X<em> = 51 g</em>
  • 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):

          Z-score=(x-\mu)/\sigma

          0.675=(51g-50.1g)/\sigma

          \sigma =0.9g/0.675=1.33g

6 0
3 years ago
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