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2 years ago

Consider the equation below (in screenshot)

Mathematics

1 answer:

frozen [14]2 years ago

7 0

The approximate solution of the equation shown in the picture is x ≈ 39 / 8 (Right choice: B).

<h3>How to find an approximate solution of a one-variable equation</h3>

The solution of the equation is between x = 4 and x = 5. Now we begin by evaluating each side of the expression (f(x) = x² - 5 · x + 4, g(x) = 2 / (x - 1)) at the average of x = 4 and x = 5.

x = (4 + 5) / 2

x = 4.5

f(4.5) = 4.5² - 5 · 4.5 + 1

f(4.5) = - 5 / 4

g(4.5) = 2 / (4.5 - 1)

g(4.5) = 4 / 7

The solution of the equation is between x = 4.5 and x = 5, then we evaluate at the average:

x = (4.5 + 5) / 2

x = 4.75

f(4.75) = 4.75² - 5 · 4.75 + 1

f(4.75) = - 3 / 16

g(4.75) = 2 / (4.75 - 1)

g(4.75) = 8 / 15

The solution of the equation is between x = 4.75 and x = 5, then we evaluate at the average:

x = (4.75 + 5) / 2

x = 4.875

f(4.875) = 4.875² - 5 · 4.875 + 1

f(4.875) = 25 / 64

g(4.875) = 2 / (4.875 - 1)

g(4.875) = 16 / 31

The approximate solution of the equation shown in the picture is x ≈ 39 / 8 (Right choice: B).

To learn more on successive approximations: brainly.com/question/27191494

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

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

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

7 0

3 years ago

If y = 3x^3 - 2x and dx/dt equals 3, find dy/dt when x = -2. Give only the numerical answer. For example, if dy, dt = 4, type on

Nadya [2.5K]

Answer:

$\frac{dy}{dt} = 102$

Step-by-step explanation:

step(i)

Given function y = 3 x³ - 2 x ... (i)

Differentiating equation (i) with respective to 'x'

$\frac{dy}{dx} = 3 (3 x^{2} ) - 2(1)$

Given x = -2

$(\frac{dy}{dx} ) x_{=-2} = 3 (3 (-2)^{2} ) - 2(1)$

$(\frac{dy}{dx} ) x_{=-2} = 36 -2 =34$

Step(ii):-

we know that

$\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

Given $\frac{dx}{dt} = 3 and \frac{dy}{dx} = 34$

$\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

substitute values $\frac{dx}{dt} = 3 and \frac{dy}{dx} = 34$

⇒ $34 = \frac{\frac{dy}{dt} }{3 }$

cross multiplication , we get

$\frac{dy}{dt} = 34( 3 ) = 102$

Final answer:-

$\frac{dy}{dt} = 34( 3 ) = 102$

8 0

3 years ago

7 less than k is no less than 70.

77julia77 [94]

Answer:

k-7 $\geq$ 70

Step-by-step explanation: it says no less than 70 implying it could be 70 or above. Greater than or equal to seventy is the correct answer.

4 0

2 years ago