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3 years ago

15

The correlation analysis assumes that the measurements have a bivariate normal distribution in the population. Select all of the

features that define a bivariate normal distribution.
A) A cloud of points that is funnel shaped (wider at one end than the other)
B) A relationship between X and Y that is not linear
C) The frequency distributions of X and Y separately are normal
D) Either X or Y has a decidedly skewed distribution
E) The presence of outliers Bell-shaped probability distribution in two dimensions rather than one
D) A relationship between X and Y that is linear

1 answer:

hoa [83]3 years ago

8 0

Answer:

Bruh i got correlation hw that i posted on here too and nobody helped lol.

Step-by-step explanation:

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If there are 12 girls and 23 boys,

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Answer:

12:23

12 to 23

12/23

Step-by-step explanation:

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3 years ago

Find AB. Round to the nearest tenth if necessary.<br>

shusha [124]

Answer:

I think the right answer would be 7

Step-by-step explanation:

We got-ta use the equation: CA * BA = DA²

CA = 15 + BA

DA² = 8² = 64

=> (15 + BA) BA = 64

Subtract 15 from each side:

BA(BA) = 49

BA² = 49

Now take the square root of each side

BA = 7

Hope this helps!

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3 years ago

8.21 in a mixed number

vfiekz [6]

It's really simple!

as an improper fraction it would look like this:

when changing it to a mixed fraction:
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the mixed number will look like this:
8 21/100
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3 years ago

Ch statement is true about the graphs of the two lines y=-6 and x =

jarptica [38.1K]

The slope answer for this question is 46% rad

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2 years ago

This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,

stich3 [128]

Answer:

a) The 4th degree , 5th degree and sixth degree polynomials

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

Step-by-step explanation:

Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

f(x) = log(1+x) ……(1)

using formula $\frac{d}{dx} logx =\frac{1}{x}$

Differentiating Equation(1) with respective to 'x' we get

$f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)$

$f^{l} (x) = \frac{1}{1+x} (1)$ ….(2)

At x= 0

$f^{l} (0) = \frac{1}{1+0} (1)= 1$

using formula $\frac{d}{dx} x^{n-1} =nx^{n-1}$

Again Differentiating Equation(2) with respective to 'x' we get

$f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))$

$f^{ll} (x) = \frac{-1}{(1+x)^2} (1)$ ….(3)

At x=0

$f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1$

Again Differentiating Equation(3) with respective to 'x' we get

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))$

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}$ ….(4)

At x=0

$f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)$

$f^{lll} (0) = 2$

Again Differentiating Equation(4) with respective to 'x' we get

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))$

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$ ....(5)

$f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}$

$f^{lV} (0) = -6$

Again Differentiating Equation(5) with respective to 'x' we get

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$ .....(6)

At x=0

$f^{V} (x) = 24$

Again Differentiating Equation(6) with respective to 'x' we get

$f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

and so on...

The nth term is

$f^{n} (x) = = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}$

Step :-2

Taylors theorem expansion of f(x) is

$f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)$

At x=a =0

$f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

Substitute all values , we get

$f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

On simplification we get

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

4 0

3 years ago