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3 years ago

Which statements describe the sequence -3,5,-7,9,11...? Check all that apply

Mathematics

1 answer:

inessss [21]3 years ago

8 0

Answer:

number series puzzele??

Step-by-step explanation:

You might be interested in

What is -15+v over -3=21

Snezhnost [94]

Answer:

V = -48

Step-by-step explanation:

((-15 + v) / -3) = 21

(-15 + v) = -63

v = -63 + 15

v = -48

4 0

3 years ago

One letter tile is redrawn form the bag abd replaced 300 times. Predict how many times a consonant will not be picked.

algol13

239 times i think, hope this helped

6 0

2 years ago

A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were w

lana66690 [7]

Answer: 0.3679

Step-by-step explanation:

The cumulative distribution function for exponential distribution is :-

$P(x)=1-e^{-\lambda x}$ , where $\lambda$ is the mean of the distribution.

Given : A mean length of waiting time equal $2.8$ minutes

i.e. Mean number of calls answered in a minute $\lambda=\dfrac{1}{2.8}$

Now, the proportion of callers is put on hold longer than 2.8 minutes is given by :_

$P(x>2.8)=1-P(x$

Hence, the proportion of callers is put on hold longer than 2.8 minutes = 0.3679

8 0

3 years 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 $U\sim N(n,1)$ :