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3 years ago

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Graph a system of equations to approximate the value of x, the rate of depreciation. Give your answer as a percent. about 4% abo

ut 13% about 1.2% cannot be determined

2 answers:

muminat3 years ago

9 0

Answer:about 13%

Step-by-step explanation:

Dovator [93]3 years ago

7 0

Answer: about 13% (B)

Step-by-step explanation:

You might be interested in

A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us

gregori [183]

Answer and Step-by-step explanation: The null and alternative hypothesis for this test are:

$H_{0}: s_{1}^{2} = s_{2}^{2}$

$H_{a}: s_{1}^{2} > s_{2}^{2}$

To test it, use F-test statistics and compare variances of each treatment.

Calculate F-value:

$F=\frac{s^{2}_{1}}{s^{2}_{2}}$

$F=\frac{1.26^{2}}{0.93^{2}}$

$F=\frac{1.5876}{0.8649}$

F = 1.8356

The <u>critical value of F</u> is given by a F-distribution table with:

degree of freedom (row): 20 - 1 = 19

degree of freedom (column): 20 - 1 = 19

And a significance level: α = 0.05

$F_{critical}$ = 2.2341

Comparing both values of F:

1.856 < 2.2341

i.e. F-value calculated is less than F-value of the table.

Therefore, failed to reject $H_{0}$ , meaning there is <u>no sufficient data to support the claim</u> that sham treatment have pain reductions which vary more than for those using magnets treatment.

4 0

3 years ago

Taylor Series Questions!

riadik2000 [5.3K]

5.
$f(x)=\sin x\implies f(\pi)=0$
$f'(x)=\cos x\implies f'(\pi)=-1$
$f''(x)=-\sin x\implies f''(\pi)=0$
$f'''(x)=-\cos x\implies f'''(\pi)=1$

Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by

$f(x)=-(x-\pi)+\dfrac{(x-\pi)^3}{3!}-\dfrac{(x-\pi)^5}{5!}+\cdots$
$f(x)=\displaystyle\sum_{n\ge0}\frac{(-1)^{n-1}(x-\pi)^{2n+1}}{(2n+1)!}$

Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at $x=0$ , not $x=\pi$ , and so the sign on each derivative at $x=\pi$ is opposite of what it should be. I'm sure you can figure out the radius of convergence from here.

- - -

6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in $x$ as a polynomial in $x-2$ .

$f(x)=x^6-x^4+2\implies f(2)=50$
$f'(x)=6x^5-4x^3\implies f'(2)=160$
$f''(x)=30x^4-12x^2\implies f''(2)=432$
$f'''(x)=120x^3-24x\implies f'''(2)=912$
$f^{(4)}(x)=360x^2-24\implies f^{(4)}(2)=1416$
$f^{(5)}(x)=720x\implies f^{(5)}(2)=1440$
$f^{(6)}(x)=720\implies f^{(6)}(2)=720$
$f^{(n\ge7)}(x)=0\implies f^{(n\ge7)}(2)=0$

$\implies f(x)=50+160(x-2)+216(x-2)^2+152(x-2)^3+59(x-2)^4+12(x-2)^5+(x-2)^6$

If you expand this, you will end up with $f(x)$ again, so the Taylor series must converge everywhere.

I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:

$x^6-x^4+2=(x-2)^6+a_5(x-2)^5+a_4(x-2)^4+a_3(x-2)^3+a_2(x-2)^2+a_1(x-2)+a_0$

You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.

- - -

7. It would help to write the square root as a rational power first:

$f(x)=\sqrt x=x^{1/2}\implies f(4)=2$
$f'(x)=\dfrac{(-1)^0}{2^1}x^{-1/2}\implies f'(4)=\dfrac1{2^2}$
$f''(x)=\dfrac{(-1)^1}{2^2}x^{-3/2}\implies f''(4)=-\dfrac1{2^5}$
$f'''(x)=\dfrac{(-1)^2(1\times3)}{2^3}x^{-5/2}\implies f'''(4)=\dfrac3{2^8}$
$f^{(4)}(x)=\dfrac{(-1)^3(1\times3\times5)}{2^4}x^{-7/2}\implies f^{(4)}(4)=-\dfrac{15}{2^{11}}$
$f^{(5)}(x)=\dfrac{(-1)^4(1\times3\times5\times7)}{2^5}x^{-9/2}\implies f^{(5)}(4)=\dfrac{105}{2^{14}}$

The pattern should be fairly easy to see.

$f(x)=2+\dfrac{x-4}{2^2}-\dfrac{(x-4)^2}{2^5\times2!}+\dfrac{3(x-4)^3}{2^8\times3!}-\dfrac{15(x-4)^4}{2^{11}\times4!}+\cdots$
$f(x)=2+\displaystyle\sum_{n\ge1}\dfrac{(-1)^n(-1\times1\times3\times5\times\cdots\times(2n-3)}{2^{3n-1}n!}(x-4)^n$

By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(-1\times\cdots\times(2n-3)\times(2n-1))(x-4)^{n+1}}{2^{3n+2}(n+1)!}}{\dfrac{(-1)^n(-1\times\cdots\tiems(2n-3))(x-4)^n}{2^{3n-1}n!}}\right|$
$\implies\displaystyle\frac{|x-4|}8\lim_{n\to\infty}\frac{2n-1}{n+1}=\frac{|x-4|}4$
$\implies |x-4|$

so that the ROC is 4.

- - -

10. Without going into much detail, you should have as your Taylor polynomial

$\sin x\approx T_4(x)=\dfrac12+\dfrac{\sqrt3}2\left(x-\dfrac\pi6\right)-\dfrac14\left(x-\dfrac\pi6\right)^2-\dfrac1{4\sqrt3}\left(x-\dfrac\pi6\right)^3+\dfrac1{48}\left(x-\dfrac\pi6\right)^4$

Taylor's inequality then asserts that the error of approximation on the interval $0\le x\le\dfrac\pi3$ is given by

$|\sin x-T_4(x)|=|R_4(x)|\le\dfrac{M\left|x-\frac\pi6\right|^5}{5!}$

where $M$ satisfies $|f^{(5)}(x)|\le M$ on the interval.

We know that $(\sin x)^{(5)}=\cos x$ is bounded between -1 and 1, so we know $M=1$ will suffice. Over the given interval, we have $\left|x-\dfrac\pi6\right|\le\dfrac\pi6$ , so the remainder will be bounded above by

$|R_4(x)|\le\dfrac{1\times\left(\frac\pi6\right)^5}{5!}=\dfrac{\pi^5}{933120}\approx0.000328$

which is to say, over the interval $0\le x\le\dfrac\pi3$ , the fourth degree Taylor polynomial approximates the value of $\sin x$ near $x=\dfrac\pi6$ to within 0.000328.

7 0

4 years ago

Mrs johnson grows herbs in square plots Her basil plot measures. 5. 9 yd on each side. A. Find the total area of the basil plot.

UNO [17]

Answer:

Area = 34. 81 yd^2

Perimeter = 86.6 feets

Step-by-step explanation:

Given the following :

Shape of Mrs. Johnson's plot = square

All sides of a square are of equal length

Measure of each side = 5.9 yd

A.) Area of plot

Area of plot = Area of a square

Area of a square(A) = a^2

Where a = side length

A = 5.9^2

A = 34.81 yd^2

B) Perimeter of Basil plot = Perimeter of a square

Converting yard to feet

1 yard = 3feets

Therefore,

5.9 yards = (3 * 5.9) = 17.7 feets

Fence is 2ft from the garden on each side,

Length of fence on each side = (2 + 17.7 + 2) Feets = 21.7 Feets

Perimeter of a square (P) = 4a

Where a = side length

P = 4 × 21.7

P = 86.6 feets

6 0

3 years ago

Kitchen goods are on sale for 30% off their regular price. In addition,all goods are subject to 6% tax and a $6 shipping charge.

RSB [31]

Blender total cost (c) = 1.06(0.7c)+6. Here, with c = $65,

Blender total cost ($65) = 1.06(0.7)($65) + 6 = $48.23

5 0

3 years ago

Solve using the histograms. How many more cantaloupes weigh 8 lb or less than watermelons that weigh 8 lb or less? A. 2

nekit [7.7K]

The correct answer is D,16

7 0

3 years ago