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zubka84 [21]
3 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]3 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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Remark

This may be a little late, but it is well worth finding all the answers. It is a perfect test question.

You should solve the angles in this order 2 5 1 4 3. This is not the only order you could use.

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<180 = <2 + 148                     <2 and 148 are supplementary

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1) All triangles have 180 degrees.

2) Diagonal bisect each other and are equal.

3) The two sides (one of which is opposite angle 5 are equal.

x + x + 148 = 180            All triangles have 180 degrees

                                      All rectangles have diagonals that are bisected and the bisected parts are equal.

2x + 148 = 180                 Subtract 148

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Angle 4 and <5 are Complementary. They  make up 90 degrees.

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x + x + 32 = 180

2x + 32 = 180                     Subtract 32 from both sides

2x = 180-32

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x = 148/2

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