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muminat
3 years ago
7

Provide an appropriate response. In order to efficiently bid on a contract, a contractor wants to be 95% confident that his erro

r is less than two hours in estimating the average time it takes to install tile flooring. Previous contracts indicate that the standard deviation is 4.5 hours. How large a sample must be selected?
a.19
b.4
c.5
d.20
Mathematics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

d. 20

Step-by-step explanation:

Standard deviation is 4.5

Margin error for the problem is 2 hours

Probability 95%, that means that the siginficance level α is 1 – p

α = 1 – 0.95 = 0.05

margin of error (ME) can be defined as follows

ME = Z(α/2) * standard deviation/ √n

Where n is the sample size

Z(0.05/2) = Z(0.025)

Using a z table Z = 1.96

Now, replacing in the equation and find n

2 = 1.96 * 4.5/ √n

2 = 8.82/√n

√n = 8.82/2

√n = 4.41

n = 4.41^2

n = 19.44 ≈ 20

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9514 1404 393

Answer:

  10.49

Step-by-step explanation:

Since we know 110 = 10² +10, we can make a first approximation to the root as ...

  √10 ≈ 10 +10/21 . . . . . where 21 = 1 + 2×integer portion of root

This is a little outside the desired approximation accuracy, so we need to refine the estimate. There are a couple of simple ways to do this.

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An approximation of √110 accurate to hundredths is 10.49.

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The other simple way to refine the root estimate is to carry the continued fraction approximation to one more level.

For n = s² +r, the first approximation is ...

  √n = s +r/(2s+1)

An iterated approximation is ...

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The adds 's' to the approximate root to make the new fraction denominator.

For this root, the refined approximation is ...

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_____

<em>Additional comment</em>

Any square root can be represented as a repeating continued fraction.

  \displaystyle\sqrt{n}=\sqrt{s^2+r}\approx s+\cfrac{r}{2s+\cfrac{r}{2s+\dots}}

If "f" represents the fractional part of the root, it can be refined by the iteration ...

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The above continued fraction iteration <em>adds</em> 1+ good decimal places to the root with each iteration. The Babylonian method described above <em>doubles</em> the number of good decimal places with each iteration. It very quickly converges to a root limited only by the precision available in your calculator.

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