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eimsori [14]
4 years ago
15

The length of a rectangular mural is 2ft. more than three times the width the area is 165 sq. ft. Find the Width And Length

Mathematics
2 answers:
olga nikolaevna [1]4 years ago
5 0
We know that the formula for the area of a rectangle is: 

A=lw 

We know that the area is 165. Now we need to solve for the dimensions. 

Let the width of the rectangle be "x"

Then the length of the rectangle is 3x+2 

Substitute. 

(3x+2)(x)=156 

Multiply the terms. 

3x^2+2x=156 

Bring the 156 over. 

3x^2+2x-156=0 

We have to use the quadratic formula to solve this now. Let us restate it: 

x= \frac{-b± \sqrt{b^2-4ac} }{2a}

Now I'm going to fast forward this because all the rest is boring stuff and substitution. The answer is: 

\frac{ \sqrt{469}-1 }{3}

Since we cannot have a negative answer, we must cancel out the other answer, which I did not include. 

Hope this helped! :) 

~Cam943, Junior Moderator
iris [78.8K]4 years ago
3 0
L×b=area
l=2+3b
2+3b ×b=165
3b²=163
b²=54.3
b=√54.3
b=7.37 or 7.4=7
b=7ft
l=2+3(7.4)
l=2+ 22.2
l=24.2=24
l=24ft
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  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

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Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

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Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

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