All categories

3 years ago

Solve for X. x2= -63 Write your answer in simplified, rationalized form.

Mathematics

1 answer:

mote1985 [20]3 years ago

6 0

this is the answer to your equation

You might be interested in

Discuss the possible reasons for the differences in price in Leomande. Discuss ways to find unit prices using proportions. The l

Maru [420]

There are many reasons why there are differences in price in Lemonade.
The first is the volume or amount of the Lemonade. Obviously, the greater the amount of the Lemonade, the higher is its price.
The price of the lemonade is given, the unit price then is
$4.99/95 ounces = $0.053/ounce
Another reason is the difference in the ingredients. Some Lemonade might have additional ingredients that could make the price higher.
The difference in materials costs in different areas may also be the reason.

6 0

3 years ago

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

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

7 0

3 years ago

A recipe calls for 1 cup of milk for every 4 cups of flour. Write the linear equation that describes this relationship.

yan [13]

The correct answer is y=4x

8 0

3 years ago

4. Julie saw 3 eagles each day she went bird-watching. How<br> many eagles did Julie see in 6 days?

Umnica [9.8K]

9514 1404 393

Answer:

18 eagles

Step-by-step explanation:

Assuming Julie went bird-watching on each of those 6 days, she saw a total of ...

(3 eagles/day)(6 day) = 3·6 eagles = 18 eagles

6 0

3 years ago

What is the value of X?

sleet_krkn [62]

Answer:

11/3 = 3 2/3 ≈ 3.66667

Step-by-step explanation:

The external angle is half the difference of the intercepted arcs:

3x = (1/2)((9x +25) -36)

6x = 9x -11

11 = 3x . . . . . . . add 11-6x

11/3 = x

_____

The attached figure is drawn to scale.

5 0

3 years ago