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olga_2 [115]
2 years ago
9

8. The travel members are voting for the American city they will visit next semester: New York (N), San Francisco (S), or Chicag

o (C). Their votes are summarized in the preference table.
---Which city is selected using the plurality-with-elimination method?

-PLEASE show work (Just look at the picture for number 8)

Mathematics
2 answers:
fiasKO [112]2 years ago
5 0

9514 1404 393

Answer:

  • plurality: San Francisco
  • plurality with elimination: New York

Step-by-step explanation:

Using the plurality method, the city with the most first-place votes wins the selection. Here, that is San Francisco, with 16 votes. Behind that is New York with 14 first-place votes.

If elimination is used, the selection with the fewest first-place votes (Chicago) is eliminated from the table, and all rows below are shifted up. Now, the first place row has 16 votes for San Francisco and 18 votes for New York.

San Francisco wins the plurality election.

New York wins the plurality-with-elimination election.

babunello [35]2 years ago
5 0

Answer:

New York (N)

Step-by-step explanation:

First S has 16 votes N has 8 + 6 = 14 votes and C has 4 votes as first choice

so we first eliminate C from the table

Next now we have 16 votes for S and  8 + 6 + 4 = 18 votes for N

So new York is selected

refer the picture below

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Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

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(1)           (12, 18, 27, ...)

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r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

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a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

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a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

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