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zubka84 [21]
2 years ago
10

Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o

f p.
Mathematics
1 answer:
goblinko [34]2 years ago
8 0

Answer:

[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval  [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

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y = amount of the 40% solution

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\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}

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