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3 years ago

5

Check my answer please?

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Hello,

In my old memories,
p(μ-2σ<=X<=μ+2σ)=95%
p(μ-3σ<=X<=μ+3σ)=99%

1000-2*200=1000-400=600
1000+2*200=1000+400=1400

Answer A

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Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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Use long division to divide x^3+x^2-2x+14 by x+3

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First, $x^3=x^2\cdot x$ , and if we multiply $x+3$ by $x^2$ we get

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Subtracting this from the numerator gives a remainder of

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Next, $-2x^2=-2x\cdot x$ , and if we multiply $x+3$ by $-2x$ we have

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and subtracting this from the previous remainder, we end up with a new remainder of

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Next, $4x=4\cdot x$ , and if we multiply $x+3$ by $4$ we get

$4x+12$

Subtracting from the previous remainder, we get a new remainder of

$2$

which contains no more factors of $x$ , so we're done.

So,

$\dfrac{x^3+x^2-2x+14}{x+3}=x^2-2x+4+\dfrac2{x+3}$

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