Answer:
North - East is the answer
Explanation:
No A
Slavery in the Chesapeake region began in 1619, when a Dutch trading vessel carrying 20 African men entered Jamestown, Virginia. The slave trade expanded in the following years. Between 1700 and 1770, the region's slave population grew from 13,000 to 250,000. By the beginning of the Revolutionary War in 1775, Black people made up nearly one-third of the region's population.
In the 1800s, the Chesapeake region became a focal point of the national controversy surrounding slavery because it was in the unique position of spanning free, border and slave states:
“Free states,” which did not support slavery, made up the northern portion of the region.
“Slave states” encompassed the southern portion of the region.
“Border states” allowed slavery but were allied with the free states, further complicated the region's politics.
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):


