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alexdok [17]
3 years ago
11

how did the colony of Plymouth and Massachusetts Bay Colony contribute to the origins of American independence and government ?

Advanced Placement (AP)
1 answer:
Gre4nikov [31]3 years ago
6 0
I think that both of those colonies were formed by Puritans. The Puritans, while not technically kicked out of England, were certainly very unpopular because they had been very closely tied up in the English Civil War shortly before that. 
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The company A produces products a, and the company B produces products b. It is known that 3 percent of companies that produce p
slava [35]

Answer:

probability = 0.183 %

Explanation:

given data

produce products = 3 %

probability for producing products b = 6.1

solution

Both companies produce different products and the likelihood of bankruptcy varies depending on the product produced. So, the bankruptcy potential of A and B companies is independent.

we will  multiply the probability of each company's bankruptcy and that will be

probability = P(A=bankrupt) ×  P(B=bankrupt)

probability = 3% × 6.1%

probability = 0.183 %

4 0
3 years ago
Which institution initiates the process of socialization in a child's life? A. School B. Society C. Community D. Family
qaws [65]
I think it's letter:
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3 years ago
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Mrrafil [7]

Answer:

Solomon Asch was a pioneering social psychologist who is perhaps best remembered for his research on the psychology of conformity. 1  Asch took a Gestalt approach to the study of social behavior, suggesting that social acts needed to be viewed in terms of their setting.

Explanation:

4 0
3 years ago
In which situation has the precautionary principle been most strictly applied? (3 points)
Cloud [144]
Answer: The last option.

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8 0
2 years ago
AP Calculus AB - Differentiating using the Product Rule
faltersainse [42]

Answer:

  h'(t) = (√t)(1 -5t²)/(2t)

Explanation:

First of all, your product rule needs to be written correctly.

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You have ...

h(t)=\sqrt{t}(1-t^2)\\\\f(t)=\sqrt{t} \qquad f'(t)=\dfrac{1}{2\sqrt{t}}=\dfrac{\sqrt{t}}{2t}\\\\g(t)=1-t^2 \qquad g'(t)=-2t\\\\h'(t)=f'(t)g(t)+f(t)g'(t)=\dfrac{\sqrt{t}}{2t}(1-t^2)+\sqrt{t}(-2t)\\\\h'(t)=\dfrac{\sqrt{t}(1-t^2-4t^2)}{2t}=\dfrac{\sqrt{t}(1-5t^2)}{2t}

4 0
3 years ago
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