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notsponge [240]
2 years ago
14

Walter and Trina are planning a campus event focused on equity and inclusion. Several students would like to help with the event

by delivering speeches. Which of the following speeches would NOT be aligned with the topics of equity and inclusion?
Equal access and opportunities in fine arts


The importance of untimed tests for all students


Equality is not equity


Diverse voices in the college curriculum
SAT
1 answer:
Shkiper50 [21]2 years ago
7 0

The following speech is not necessary aligned with the topics of equity and inclusion that are planned for the event:

  • The importance of untimed tests for all students.

The speech can be considered related to equity because it is about all students, but the topic it talks about is not a very important problem in relation to inclusion and equity. However, all the other options are related.

Inclusion means that everyone is included, no one is left out. Equity means that everything is the same for every one, there are no preferences, everything is fair for everyone.

Check more information here brainly.com/question/23822990?referrer=searchResults

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul
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Sampling distribution involves the proportions of a data element in a given sample.

  • <em>The proportion of Good TV set is 0.67</em>
  • <em>The number of ways of selecting 5 from 6 TV sets is 6</em>
  • <em>The number of ways of selecting 4 from 6 TV sets is 15</em>

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Given

n = 6

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

Good = 4

So, the proportion (p) that are good are:

p = \frac{Good}{n}

p = \frac{4}{6}

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<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

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So, we have:

^6C_5 = \frac{6!}{(6 - 5)!5!}

^6C_5 = \frac{6!}{1!5!}

^6C_5 = \frac{6 \times 5!}{1 \times 5!}

^6C_5 = \frac{6}{1}

^6C_5 = 6

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<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

r = 4

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^6C_4 = \frac{6!}{(6 - 4)!4!}

^6C_4 = \frac{6!}{2!4!}

^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}

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^6C_4 = 15

So, the table is as follows:

\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

p = \frac{Good}{n}

<u>(d) The sampling distribution</u>

In (a), we have:

p = 0.67 --- proportion of Good TV

The sampling error is calculated as follows:

SE_n = |p - p_n|

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\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]

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