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3 years ago

The standard deviations of four data sets are shown in the table below. Which

Mathematics

1 answer:

seropon [69]3 years ago

4 0

You are correct. The higher the standard deviation is, the more spread out the data set will be. Nice work.

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Find the slope from the table using the slope formula

larisa86 [58]

Answer:

it false

Step-by-step explanation:

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3 years ago

Find the slope of a line perpendicular to <br><br> Y= -x -1

ad-work [718]

<h2>Greetings!</h2>

Answer:

The slope is 1.

Step-by-step explanation:

First, we must find the slope of the current equation.

This is the number in front of the x.

Seeing as this is -x, or -1x, the slope of this line is -1

When finding the slope of a line perpendicular, you need to find the $\frac{-1}{slope}$

So, in this case it is:

$\frac{-1}{-1}$

The minuses cancel out, leaving with 1 over 1 or 1/

So the slope of the line perpendicular is 1!

<h2>Hope this helps!</h2>

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3 years ago

Which choice is equivalent to the product below? √(6/8)•√(6/18) A. 6/12 B. 1/4 C. 7/12 D 3/4 E. 7/1

adoni [48]

Answer:6/12

Step-by-step explanation: ap ex

3 0

2 years ago

Read 2 more answers

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago

Find the length of BC.<br><br><br><br>Explain how you got it, please!<br>Thanks!

Vinvika [58]

Answer:

BC = 30.73

Here,

$\sf \frac{AB}{BQ} = \frac{CD}{DQ}$

so first solve for QD

$\sf \hookrightarrow \frac{32}{15} = \frac{19.2}{DQ}$

$\sf \hookrightarrow 32(DQ)} =19.2(15)$

$\sf \hookrightarrow 32(DQ)} =288$

$\sf \hookrightarrow DQ =9$

Hence, QD = 9

Now! <u>using Pythagoras theorem,</u>

CD² + BD² = BC²

19.2² + (9+15)² = BC²

BC = √368.64+576

BC = 30.73499634

BC = 30.73 ( rounded to nearest hundredth )

4 0

1 year ago

Read 2 more answers