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2 years ago

The equation for the vertical line that goes through point V (−5, 17).

Mathematics

1 answer:

valentinak56 [21]2 years ago

4 0

Answer:

The answer to the above question is x = -5

You might be interested in

16 increased by twice a number is -24

wlad13 [49]

X-the number

16 + 2x = -24 |subtract 16 from both sides

2x = -40 |divide both sides by 2

x = -20

Answer: The number is -24

8 0

3 years ago

What is the simplified value of the expression below?

Marta_Voda [28]

Answer:

A. 18.25

Step-by-step explanation:

After you do the multiplications, the problem is

... (56 +90)/8 = 146/8 = 18.25

_____

The division bar is a grouping symbol, equivalent to parentheses around both numerator and denominator. You must evaluate the numerator before you can divide by the denominator, which also must be evaluated before that division.

5 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

Please help me solve this equation

elena-s [515]

Answer:

m<U = 38degrees

Step-by-step explanation:

From the given diagram, <B = <U since both triangles are similar, hence;

Hence;

2y+2 = 3y-16

2y - 3y = -16 - 2

-y = -18

y = 18

Get m<U

m<U = 3y-16

m<U = 3(18) -16

m<U = 54 - 16

m<U = 38degrees

3 0

3 years ago

Lisa took out a loan for 5 months and was charged simple interest at an annual rate of 4.8%.

denis-greek [22]

Answer:

Lisa borrowed $8,500

Step-by-step explanation:

Simple Interest

Occurs when the interest is calculated on the original principal of a loan only.

Unlike compound interest where the interest earned in the compounding periods is added to the old principal, simple interest only considers the principal to calculate the interest.

The interest earned is calculated as follows:

I=Prt

Where:

I = Interest

P = initial principal balance

r = interest rate

t = time

Lisa took out a loan for t=5 months and was charged simple interest at an annual rate of r=4.8% = 0.048. She paid interest for I=$170.

We need to convert the time to years (there are 12 months per year):

t = 5 /12 years.

The formula must be solved for P:

$\displaystyle P=\frac{I}{rt}$

Substituting:

$\displaystyle P=\frac{170}{0.048*5/12}$

$\displaystyle P=\frac{170}{0.02}=8,500$

Lisa borrowed $8,500

8 0

3 years ago