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2 years ago

5

7=x+3. What is being done to the variable?

1 answer:

creativ13 [48]2 years ago

3 0

Step-by-step explanation: If you take a look at the right side of the equation, a 3 is being added to our variable <em>x</em>.

If you were going to solve this equation, you must subtract 3 from the right side and the left side to isolate our variable.

You would end up with 4 = x.

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which of the following is a non depository financial institution? a.Retail Bank b.commercial bank c.life insurance d.credit unio

Artyom0805 [142]

Answer: it was online bank

Step-by-step explanation:

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2 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

2 years ago

Consider the functions f(x)=(ax^2-b)/(x^2-4) and g(x)=(x^2-cx+d). if the limit as x approaches 3 equals 6, what is the relations

Dmitry_Shevchenko [17]

$6=\dfrac{a\cdot3^2-b}{3^2-4}\\\\
6=\dfrac{9a-b}{5}\\\\\
9a-b=30\\\\
\boxed{b=9a-30}$

6 0

2 years ago

Please help answer these questions. My teacher said they were really easy but I just don't understand. Will mark brainliest !!!

kodGreya [7K]

Answer:

1. A = 59

2. A = 43

Step-by-step explanation:

If we have a right triangle we can use sin, cos and tan.

sin = opp/ hypotenuse

cos= adjacent/ hypotenuse

tan = opposite/ adjacent

For the first problem, we know the opposite and adjacent sides to angle A

tan A = opposite/ adjacent

tan A = 8.8 / 5.2

Take the inverse of each side

tan ^-1 tan A = tan ^-1 (8.8/5.2)

A = 59.42077313

To the nearest degree

A = 59 degrees

For the second problem, we know the adjacent side and the hypotenuse to angle A

cos A = adjacent/hypotenuse

cos A = 15.3/21

Take the inverse of each side

cos ^-1 cos A = cos ^-1 (15.3/21)

A = 43.23323481

To the nearest degree

A = 43 degrees

6 0

3 years ago