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3 years ago

Find the average rate of change of the function from x=1 to x=2. f(x)=−14/x^2

Mathematics

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

The average rate of change of the function from x=1 to x=2 will be: 10.5

Step-by-step explanation:

Given the function

$f\left(x\right)=-\frac{14}{x^2}$

at x₁ = 1,

f(x₁) = f(1) = -14/(1)² = -14/1 = -14

at x₂ = 2,

f(x₂) = f(2) = -14/(2)² = -14/(4) = -3.5

Using the formula to determine the average rate of change at which the total cost increases will be:

Average rate of change = [f(x₂) - f(x₁)] / [ x₂ - x₁]

= [-3.5 - (-14)] / [2 - 1]

= [-3.5 + 14] / [1]

= 10.5 / 1

= 10.5

Therefore, the average rate of change of the function from x=1 to x=2 will be: 10.5

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15 a2 - 6ab- 8 a2+ 20 - 5ab - 31 + a2- ab

Ivanshal [37]

Step-by-step explanation:

Remove the parentheses: 15a²-6ab+8a²+20+5ab-31+a²-ab Combine like terms: 24a²-2ab-11

Answer: 24a²-2ab-11

4 0

2 years ago

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Help please on this math problem: The prices of backpacks at a store are 22, 16, 39, 35, 19, 34, 20, and 26. Find the Mean Absol

Ivanshal [37]

Answer: 7.22
(note: this is a result after rounding. The result before rounding was 7.21875)

========================================================

Explanation:

Given Set of Values = {22, 16, 39, 35, 19, 34, 20, 26}

Add up the values: 22+16+39+35+19+34+20+26 = 211
Divide that sum by 8 as there are 8 values: 211/8 = 26.375

The mean is 26.375

Now subtract the mean from each data value. Apply the absolute value to ensure the difference is never negative

|22 - 26.375| = 4.375
|16 - 26.375| = 10.375
|39 - 26.375| = 12.625
|35 - 26.375| = 8.625
|19 - 26.375| = 7.375
|34 - 26.375| = 7.625
|20 - 26.375| = 6.375
|26 - 26.375| = 0.375

Add up those results
4.375+10.375+12.625+8.625+7.375+7.625+6.375+0.375 = 57.75
Then divide by 8
57.75/8 = 7.21875

The mean absolute deviation of the prices is 7.21875
Rounded to two decimal places, it is 7.22
Since we're talking about money, it makes sense to round to the nearest penny.

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3 years ago

For which acute angle are the sine of the angle and cosine of the angle equal? A. 30º B. 45º C. 60º D. 75º

Vera_Pavlovna [14]

For a right triangle, the sine of an angle is the ratio of the length of the side opposite to it and the hypotenuse and the its cosine is the ratio between the side adjacent to it and the hypotenuse. From the given choices, the angle being referred to is letter B. 45°.

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2 years ago

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

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2 years ago

In the triangle shown below, what is the approximate value of x?

Verdich [7]

Answer:

the answer the answer is the answer is C

Step-by-step explanation:

21 Unity

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2 years ago

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