To determine the organic material in a dried lake bed, the percent carbon by mass is measured at two different locations. To compare the means of the two different locations, it must first be determined whether the standard deviations of the two locations are different. For each location, calculate the standard deviation and report it with two significant figures.

Answers

Answer 1

Answer:

[tex]\sigma_1 = 0.08[/tex] --- Location 1

[tex]\sigma_2 = 0.34[/tex] --- Location 2

Step-by-step explanation:

Given

See attachment for the given data

Required

The standard deviation of each location

For location 1

First, calculate the mean

[tex]\bar x_1 =\frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x_1 =\frac{30.40+30.20+30.30+30.40+30.30}{5}[/tex]

[tex]\bar x_1 =\frac{151.60}{5}[/tex]

[tex]\bar x_1 =30.32[/tex]

The standard deviation is calculated as:

[tex]\sigma_1 = \sqrt{\frac{\sum(x - \bar x_1)^2}{n-1}}[/tex]

[tex]\sigma_1 = \sqrt{\frac{(30.40 - 30.32)^2+(30.20 - 30.32)^2+(30.30 - 30.32)^2+(30.40 - 30.32)^2+(30.30 - 30.32)^2}{5-1}}[/tex]

[tex]\sigma_1 = \sqrt{\frac{0.028}{4}}[/tex]

[tex]\sigma_1 = \sqrt{0.007}[/tex]

[tex]\sigma_1 = 0.08[/tex]

For location 2

First, calculate the mean

[tex]\bar x_2 =\frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x_2 =\frac{30.10+30.90+30.20+30.70+30.30}{5}[/tex]

[tex]\bar x_2 =\frac{152.2}{5}[/tex]

[tex]\bar x_2 =30.44[/tex]

The standard deviation is calculated as:

[tex]\sigma_2 = \sqrt{\frac{\sum(x - \bar x_2)^2}{n-1}}[/tex]

[tex]\sigma_2 = \sqrt{\frac{(30.10-30.44)^2+(30.90-30.44)^2+(30.20-30.44)^2+(30.70-30.44)^2+(30.30-30.44)^2}{5-1}}[/tex]

[tex]\sigma_2 = \sqrt{\frac{0.472}{4}}[/tex]

[tex]\sigma_2 = \sqrt{0.118}[/tex]

[tex]\sigma_2 = 0.34[/tex]

To Determine The Organic Material In A Dried Lake Bed, The Percent Carbon By Mass Is Measured At Two

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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

Find the first, second, third and fourth order Maclaurin polynomials of f(x) =

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diagram. (Sketch by hand or use software.)

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The first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

The first order Maclaurin polynomial is f(x)=xThe second order Maclaurin polynomial is also f(x)=xThe third order Maclaurin polynomial is [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]The fourth order Maclaurin polynomial is also [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]You can see the graph on the attached picture.

So let's start by finding the first order maclaurin polynomial:

f(x)=f(0)+f'(0)x

so let's find each part of the function:

f(0)=arctan(0)

f(0)=0

now, let's find the first derivative of f(x)

f(x)=arctan(x)

This is a usual derivative so there is a rule we can use here:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

so now we can find f'(0)

[tex]f'(0)=\frac{1}{(0)^{2}+1}[/tex]

f'(0)=1

So we can now complete the first order Maclaurin Polynomial:

f(x)=0+1x

which simplifies to:

f(x)=x

Now let's find the second order polynomial, for which we will need to get the second derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}[/tex]

so:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

we can rewrite this derivative as:

[tex]f'(x)=(x^{2}+1)^{-1}[/tex]

and use the chain rule to get:

[tex]f''(x)=-1(x^{2}+1)^{-2}(2x)[/tex]

which simplifies to:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

now, we can find f''(0):

[tex]f''(0)=-\frac{2(0)}{((0)^{2}+1)^{2}}[/tex]

which yields:

f''(0)=0

so now we can complete the second order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}[/tex]

which simplifies to:

f(x)=x

Now let's find the third order polynomial, for which we will need to get the third derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}[/tex]

so:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

In this case we can use the quotient rule to solve this:

Quotient rule: Whenever you have a function in the form , then it's derivative is:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

p=2x

p'=2

[tex]q=(x^{2}+1)^{2}[/tex]

[tex]q'=2(x^{2}+1)(2x)[/tex]

[tex]q'=4x(x^{2}+1)[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

[tex]f'''(x)=\frac{(2)(x^{2}+1)^{2}-(2x)(4x)(x^{2}+1)}{((x^{2}+1)^{2})^{2}}[/tex]

which simplifies to:

[tex]f'''(x)=\frac{-2x^{2}-2+8x^{2}}{(x^{2}+1)^{3}}[/tex]

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

now, we can find f'''(0):

[tex]f'''(0)=\frac{6(0)^{2}-2}{((0)^{2}+1)^{3}}[/tex]

which yields:

f'''(0)=-2

so now we can complete the third order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

Now let's find the fourth order polynomial, for which we will need to get the fourth derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}[/tex]

so:

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

In this case we can use the quotient rule to solve this:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

[tex]p=6x^{2}-2[/tex]

p'=12x

[tex]q=(x^{2}+1)^{3}[/tex]

[tex]q'=3(x^{2}+1)^{2}(2x)[/tex]

[tex]q'=6x(x^{2}+1)^{2}[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

[tex]f^{4}(x)=\frac{(12x)(x^{2}+1)^{3}-(6x^{2}-2)(6x)(x^{2}+1)^{2}}{((x^{2}+1)^{3})^{2}}[/tex]

which simplifies to:

[tex]f^{4}(x)=\frac{12x^{3}+12x-6x^{3}+12x}{(x^{2}+1)^{4}}[/tex]

[tex]f^{4}(x)=\frac{6x^{3}+24x}{(x^{2}+1)^{4}}[/tex]

now, we can find f^{4}(0):

[tex]f^{4}(x)=\frac{6(0)^{3}+24(0)}{((0)^{2}+1)^{4}}[/tex]

which yields:

[tex]f^{4}(0)=0[/tex]

so now we can complete the fourth order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}+\frac{0}{4!}x^{4}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

you can find the graph of the four polynomials in the attached picture.

So the first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

The first order Maclaurin polynomial is f(x)=xThe second order Maclaurin polynomial is also f(x)=xThe third order Maclaurin polynomial is [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]The fourth order Maclaurin polynomial is also [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

You can find further information on the following link:

https://brainly.com/question/17440012?referrer=searchResults

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