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3 years ago

If the number 3 were added to the set {1,1,3,3,4,4,5}, which central tendency would change?

Mathematics

1 answer:

ziro4ka [17]3 years ago

8 0

I think it would be the mean

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PLEASE HELP ASAP!! URGENT!!

Black_prince [1.1K]

A= 9
B= 11

Make sure to try it

4 0

3 years ago

Khan Academy Question

Mama L [17]

Answer:

Step-by-step explanation:

Since we are given line OP perpendicular to line DR, then angles PDR and ODR are right angles.

Angles PDA and ADR are complementary.

Angles ODU and UDR are complementary.

Angles ADR and UDR are given as congruent.

We can conclude that angles PDA and ODU are congruent.

By AA Similarity, triangles APD and UMD are similar.

DP/DM = PA/MU

3.75/10 = 4.5/MU

3.75MU = 10 × 4.5

MU = 12 (altitude of triangle DUO)

OD = OM + MD = 4 + 10 = 14 (base of triangle DUO)

area = base × height / 2

area = 14 × 12 / 2

area = 84

4 0

1 year ago

First convert these fractions to twelfths then order them.Write your

sergiy2304 [10]

Here’s the link to the answer

Curlylips444.com

8 0

3 years ago

What are some Examples of similar triangles or polygons that you may see in the real world?

Fofino [41]

Answer:

You can see objects, nd buildings that have those shapes, and roof tops.

Step-by-step explanation:

8 0

2 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 $U\sim N(n,1)$ :

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

so we end up with

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

3 0

2 years ago