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3 years ago

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A car averages 22 miles per gallon. If gas costs $2.89 per gallon, which of the following is closest to how much the gas would c

ost for this car to travel 1,650 typical miles?

1 answer:

Alja [10]3 years ago

4 0

You take 1,650/22=75
Then 75•2.89
So your answer is $216.75 is how much it would cost to go 1,650 miles

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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:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

<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:

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

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

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

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

$^6C_5 = 6$

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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:

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Recent data show that phenotypic variation categories such as skin color or facial form most often used to define race.

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Skin color

Height

Shapes and sizes

Based on recent data from scientific research, phenotypic variation categories such as skin color or facial form most often used to define race.

Read more on phenotypic variation here: brainly.com/question/22840871

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