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Scorpion4ik [409]
3 years ago
12

A pharmacist has an 18% alcohol solution and a 40% alcohol solution. How much of each should he mix together to make 10 liters o

f a 20% alcohol solution?
Mathematics
1 answer:
ladessa [460]3 years ago
3 0
X = amount of the 18% solution

y = amount of the 40% solution

we know the 18% solution has only 18% of alcohol, the rest is maybe water or something, now, how many liters is 18%?  well, 18% of anything is just (18/100) * anything, so, 18% of x is just (18/100) *x or 0.18x, and that's how many liters are there.

likewise, how many liters are there in the 40% solution?  well, (40/100) * y, or 0.4y, that many.

we know the mixture has to yield 10 liters at 20% alcohol, how many liters of only alcohol is that?  well, (20/100) * 10, or 2 liters.

\bf \begin{array}{lccclll}
&\stackrel{liters}{amount}&\stackrel{alcohol~\%}{quantity}&\stackrel{liters}{alcohol}\\
&------&------&------\\
\textit{18\% solution}&x&0.18&0.18x\\
\textit{40\% solution}&y&0.40&0.4y\\
------&------&------&------\\
mixture&10&0.20&2
\end{array}

\bf \begin{cases}
x+y=10\implies \boxed{y}=10-x\\
0.18x+0.4y=2\\
----------\\
0.18x+0.4\left( \boxed{10-x} \right)=2
\end{cases}
\\\\\\
0.18x-0.4x+4=2\implies -0.22x=-2
\\\\\\
x=\cfrac{-2}{-0.22}\implies x=\cfrac{100}{11}\implies x=9\frac{1}{11}

how much of the 40% solution?  well y = 10 - x
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Answer:

a

The 95% confidence interval is

   0.7811 <  p <  0.9529

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    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  

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 n =  125  \  stores  

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  60  

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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    

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Generally the margin of error is mathematically represented as  

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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    

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Generally the sample size is mathematically represented as  

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

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