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8 months ago

NEED HELP ASAP Is each line parallel, perpendicular, or neither to a line whose slope is 3/8?

Mathematics

1 answer:

Lostsunrise [7]8 months ago

4 0

Line e is parallel.

Line h is perpendicular.

Lines f and g are neither.

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The useful life of an electrical component is exponentially distributed with a mean of 1000 hours a. What is the probability the

julia-pushkina [17]

Answer:

0.865

Step-by-step explanation:

To solve for the probability we would be using the formula:

F(x) = 1 - e^−λx

Therefore, P(x > 2000) = F(x) = 1 - e^−λx

Mean = 1000 hours

λ = 1/mean = 1/1000 hours

x= 2000 hours

F(x) = 1 - e^−(1/1000)2000

F(x) = 1 - e^−2

= 0.8646647168

Therefore, the probability the the circuit will last at least 2000 hours = 0.865

3 0

2 years ago

WILL MARK BRAINLIEST ASAP PLEASE

ohaa [14]

Answer:

Step-by-step explanation:

90×(2)

= 180

each mile is equal to 90 calories

therefore 2 miles is equal to 180

8 0

1 year ago

Select the two expressions that equivalent to 4 + w + 12w

Korvikt [17]

Answer:

A and C :)

Have an amazing day!!

Please rate and mark brainliest!!

4 0

1 year ago

PLEASE HELP!

emmainna [20.7K]

Use the given recursion and starting value of $x_0 = 2.4$ to find $x_1$ :

$x_1 = \dfrac{x_0 + \frac6{x_0}}2 = \dfrac{2.4 + \frac{6}{2.4}}2 = 2.45$

Do the same for $x_2$ and $x_3$ :

$x_2 = \dfrac{x_1 + \frac6{x_1}}2 = \dfrac{2.45 + \frac6{2.45}}2 \approx 2.44949$

$x_3 = \dfrac{x_2+\frac6{x_2}}2 \approx \dfrac{2.44949 + \frac6{2.44949}}2 \approx \boxed{2.44949}$

(That's not a mistake. This just tells you that the 2nd and 3rd iterates are very close together and have at least the same first 5 digits after the decimal.)

5 0

1 year ago

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

2 years ago