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Yakvenalex [24]
2 years ago
10

Can anyone pls tell me when do we cross multiply and when do we multiply 10 * 75 im really confused

Mathematics
1 answer:
Westkost [7]2 years ago
4 0

You don't multiply liters of gasoline by distance.

They are<u> directly proportional </u>- the more gasoline you have the longer is the dictance you can go.

So it's <u>cross multiply</u>.

Another examlpe:

you are paid by hour for your work. The more hours you have worked, the more you are paid:

hours      payment

2 h            $36

5 h             x    

x=\dfrac{5\cdot36}2=90

You would multiply 10•70 if they were inversely poportional values

For example:

1. speed and time of ride by given distance {the greater the speed the less time you need for a ride}

speed          time

10 m/s          75 s

 8 m/s           x

x=\dfrac{10\cdot75}{8}=93.75\,s

2. price and kolos of apples - the higher the pice the less of kilos you can buy by given cash

price          kilos of apples

$2                8

$4                x

x=\dfrac{2\cdot8}{4}=4\,kg

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Answer:

a

The 95% confidence interval is

   0.7811 <  p <  0.9529

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

b

  Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

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Step-by-step explanation:

From the question we are told that

    The sample size is  n =  60  

    The number of stores that had more than 2 items price incorrectly is  k =  52  

   

Generally the sample proportion is mathematically represented as  

             \^ p  =  \frac{ k }{ n }

=>          \^ p  =  \frac{ 52 }{ 60 }

=>          \^ p  =  0.867

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

=>   E =  1.96  * \sqrt{\frac{ 0.867  (1- 0.867)}{60} }

=>   E =  0.0859

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.867  - 0.0859  <  p <  0.867  +  0.0859

=>    0.7811 <  p <  0.9529

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

Considering question b

Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

   

Considering question c

From the question we are told that

    The margin of error is  E = 0.05

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=> n=  [\frac{1.645 }}{0.05} ]^2 * 0.867  (1 - 0.867 )

=>   n =  125  \  stores  

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