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worty [1.4K]
3 years ago
11

: 88, 87, 89 90, 87, 85, 88, 91, 86, 86, 88, 89what's is the inerquartile range​

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
8 0
I think 88 i don’t really know
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Answer:

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Step-by-step explanation:

we have

2x^{2} +20x=-38

Divide by 2 both sides

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we know that


The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}


in this problem we have


x^{2} +10x+19=0

so


a=1\\b=10\\c=19


substitute

x=\frac{-10(+/-)\sqrt{10^{2}-4(1)(19)}}{2(1)}


x=\frac{-10(+/-)\sqrt{100-76}}{2}


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x=\frac{-10(+/-)2\sqrt{6}}{2}


x1=\frac{-10(+)2\sqrt{6}}{2}=-5+\sqrt{6}


x2=\frac{-10(-)2\sqrt{6}}{2}=-5-\sqrt{6}


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