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goldfiish [28.3K]
3 years ago
13

__________ was a major figure in the studies of social interaction.

Mathematics
1 answer:
Radda [10]3 years ago
3 0

Answer:

Erving Goffman

Step-by-step explanation:

  • <em>George Herbert Mead</em> behaviorism's theory believed that people develop self-images through interactions with other people, his four basic ideas of how the self (image and awareness) develop are: The self develop solely through social experience, social experience consist on the exchange of symbols, knowing other's intentions and understanding the role of the other results in self-awareness.
  • <em>Charles Horton Cooley</em>  also believed in how self-image is influenced by the interaction with others, especially with significant others (person whose opinions matter to us and who influence our thinking)
  • Erving Goffman proposed the social action theory, he defends that human behavior depends on personal scenarios and relationships, this means we try to handle the impressions about us that others form. He is recognized for his major contributions to sociology, a pioneer of micro-sociology, or the close examination of social interactions in everyday life.
  • <em>Jean Piaget </em>developed the cognitive theory that explains how a child constructs a mental model of the world, it consists of building blocks of knowledge, adaptation processes that enable the transition from one stage to another and stages of cognitive development.

I hope you find this information useful and interesting! Good luck!

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Step-by-step explanation:

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Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

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  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

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  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
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