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trasher [3.6K]
2 years ago
11

A student wants to verify a claim that for a given change in momentum, the average applied force decreases as the time interval

increases. The student plans to use a force sensor to produce graphs of force as a function of time for various trials of an experiment. Which of the following is the best method for the student to use to support the claim?.
SAT
1 answer:
Kipish [7]2 years ago
3 0
Here’s the answer hope this is correct

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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>

<em />

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}

p = 0.67

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

This is calculated using:

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

Where

r = 5

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

Hence, there are 6 ways

<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

So, we have:

^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!}

^6C_4 = \frac{30}{2}

^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|

So, we have:

\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]

Read more about sampling distributions at:

brainly.com/question/10554762

3 0
3 years ago
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