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3 years ago

What is the interquartile range of this data set -5,-4,-3,-2,-1,0,1,2,3,4,5,7,7,7

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

4 0

Answer:

IQR=5+2=7 because you have to find the median and the median below the median and above the median to get 5 and a negative two and 5 subtract a negative is the same as adding a positive so then you get 7.

Step-by-step explanation:

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(a + b - c )(a + b + c )

denis-greek [22]

Answer:

$a^2+b^2-c^2+2ab$

Step-by-step explanation:

$(a+b-c)(a+b+c)=a^2+ab+ac+ab+b^2+bc-ac-bc-c^2=a^2+b^2-c^2+2ab$

Hope this helps!

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3 years ago

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What is the distance between the points (–3, 4) and (–7, 4)?

uysha [10]

Answer:

$d = \sqrt{( - 7 + 3) {}^{2} + (4 - 4) {}^{2} } \\ d = \sqrt{16} \\ d = 4$

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4 years ago

The lowest temperature in January was less than−6°. Which could be the lowest temperature in January?

PIT_PIT [208]

Answer

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Step-by-step explanation:

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3 years ago

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Simplify: 18q2−45q+25/9q2−25.

Svet_ta [14]

Answer:

$\frac{(6q-5)}{(3q+5)}$

Step-by-step explanation:

Given: $\frac{18q^{2}-45q+25 }{9q^{2}-25 }$

Factorizing the numerator and the denominator, we have:

18 $q^{2}$ - 45q + 25 = 18 $q^{2}$ - 30q - 15q + 25

= (3q - 5)(6q - 5)

and

9 $q^{2}$ - 25 = (3q - 5)(3q + 5)

So that,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}$

= $\frac{(6q-5)}{(3q+5)}$

Therefore,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(6q-5)}{(3q+5)}$

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3 years ago

What is 298.8 subtracted by 4.09

adelina 88 [10]

Answer:

294.71

Step-by-step explanation:

298.8-4.09=294.71

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3 years ago