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In-s [12.5K]
3 years ago
5

Find the interquartile range (IQR) of the data in the box plot below. A horizontal boxplot titled Magnitude of earthquakes, is p

lotted along a horizontal axis marked from 0 to 10, in increments of 0.5. A left whisker extends from 2.5 to 4. The box extends from 4 to 7 and is divided into 2 parts by a vertical line segment at 6.5. The right whisker extends from 7 to 8.5. All values estimated.
Mathematics
2 answers:
tresset_1 [31]3 years ago
8 0

Answer:

IQR = 3

Step-by-step explanation:

Interquartile range, which is abbreviated as IQR, is just the difference between the upper quartile (where the box ends) and the lower quartile (where the box starts). AS it is given in the question that that box starts from 4 and ends at . The interquartile range IQR is simply:

IQR = 7 - 4 = 3

The values of data are usually clustured around some central value. Interquartile range tells us how spread out these middle values. It also tells when some values are too far from the central value. The values which are too far are also termed as outlier values

elena55 [62]3 years ago
3 0

Answer:

6.5

Step-by-step explanation:

for anybody else still wondering...

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Pani-rosa [81]

The midsegment is half the length of the base. So the base = 2*4 = 8 in.

Since it is an isosceles triangle, the other two sides are equal in length.

Perimeter = sum of two equal sides + base = 20 in.

2x equal side + 8 = 20

equal side = 12/2 = 6

Each of the equal sides = 6"


4 0
3 years ago
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What type of distribution is shown in<br> the graph?
AnnZ [28]

Answer:

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HELP?!? idk what im doing and nobody has asked this. will give brainliest answrr or whatever its called.
Illusion [34]

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Answer:

  \textbf{A.   }x=\dfrac{-4\pm2\sqrt{13}}{3}

Step-by-step explanation:

First of all, subtract 12 from both sides of the equation to put it into standard form:

  3x² +8x -12 = 0

Then identify the a, b, c of this form:

  ax² +bx +c = 0

You see that a=3, b=8, c=-12.

Now use those values in the quadratic formula for the solutions:

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-8\pm\sqrt{8^2-4(3)(-12)}}{2(3)}\\\\x=\dfrac{-8\pm\sqrt{208}}{6}=\dfrac{-8\pm4\sqrt{13}}{6}\\\\\boxed{x=\dfrac{-4\pm2\sqrt{13}}{3}}\qquad\text{factor 2 from numerator and denominator}

6 0
3 years ago
construct a right-angled triangle ABC where angle A =90 degree , BC= 4.5cm and AC= 7cm. please ans fast........ Very urgent. Pls
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Answer and Step-by-step explanation: The described right triangle is in the attachment.

As it is shown, AC is the hypotenuse and BC and AB are the sides, so use Pytagorean Theorem to find the unknown measure:

AC² = AB² + BC²

AB^{2} = AC^{2}-BC^{2}

AB =\sqrt{AC^{2}-BC^{2}}

AB =\sqrt{7^{2}-4.5^{2}}

AB =\sqrt{28.75}

AB = 5.4

Then, right triangle ABC measures:

AB = 5.4cm

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6 0
3 years ago
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Liono4ka [1.6K]

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The double angle formula for the sine is sin 2theta = 2 sin theta *cos theta.

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3 years ago
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