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3 years ago

Find the rational approximation of √15.

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

7 0

9514 1404 393

Answer:

1921/496

Step-by-step explanation:

There are many rational approximations to √15, some better than others.

A linear approximation is often used:

√15 ≈ 3 +(15-3^2)/(4^2-3^2) = 3 6/7 = 27/7

That can be refined by one iteration of the Babylonian method of determining the root:

√15 ≈ (27/7 +15/(27/7))/2 = (3 6/7 +3 8/9)/2 = 3 55/63 = 244/63

This value is equivalent to the root rounded to 4 decimal places.

Another iteration of the Babylonian method gives the approximation ...

√15 ≈ 119071/30744, equivalent to the root rounded to 9 decimal places.

The approximation 1921/496 is the best approximation that has a denominator under 1000.

_____

Additional comment

Successive rational convergents of the continued fraction approximation of √15 can be found as ...

a'/b' = (3a +15b)/(a +3b)

This method adds approximately one more good decimal place per iteration.

Successive rational approximations can be found using the Babylonian method (Newton's method iteration) as ...

a'/b' = (a² +15b²)/(2ab)

This method has "quadratic" convergence. It approximately doubles the number of good decimal places with each iteration.

You can use 3/1, 4/1, or 27/7 to begin either of these iterations.

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4 years ago

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Step-by-step explanation:

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2 years ago

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

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Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$