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masha68 [24]
3 years ago
15

Can someone help me? I’m homeschooled and I have no clue how to do a box and whisker plot. I need help ASAP so I can end this sc

hool year!!

Mathematics
1 answer:
zheka24 [161]3 years ago
7 0
<span>the ends of the box are the upper and lower quartiles, so the box spans the interquartile range.the median is marked by a vertical line inside the box.the whiskers are the two lines outside the box that extend to the highest and lowest observations.</span><span>The first step in constructing a </span>box-and-whisker plot<span> is to first find the median (Q2), the lower </span>quartile<span> (Q1) and the upper </span>quartile<span> (Q3) of a given set of data. You are now ready to find the interquartile range (IQR). ... The IQR is a very useful measurement.</span>
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Solve for x.<br><br> x=<br><br> Look at image
postnew [5]

Answer:

x = 14

Step-by-step explanation:

To find x, the following equation is generated given the available information:

\frac{(x + 7) + x)}{x + 7} = \frac{12 + 8}{12}

\frac{2x + 7}{x + 7} = \frac{20}{12}

\frac{2x + 7}{x + 7} = \frac{5}{3}

Cross multiply

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Collect like terms

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8 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
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