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Zinaida [17]
3 years ago
7

the perimeter of a rectangle field is 340 yd. the length is 30 yd longer than the width. find the dimensions

Mathematics
2 answers:
zvonat [6]3 years ago
8 0

Lets assume:

Width=x

Length=x+30

2x+2(x+30)=340

2x+2x+60=340

4x+60=340

4x=280

x=70

Plugging x in:

Width=70

Length=100

Ivahew [28]3 years ago
4 0
Let l: length, w=width
2(l+w)=340, l=w+30
l+w=170
w+30+w=170
2w=140
w=70 yd.
l=70+30=100 yd
So the length of the rectangle field is 100 yd. and the width of the rectangle field is 70 yd. Hope it help!
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Answer:

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We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation for the sampling distribution would be:

\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4

So then the answer is TRUE

Step-by-step explanation:

Let X the random variable of interest and we know that the true mean and deviation for this case are given by:

\mu = 114, \sigma = 40

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation for the sampling distribution would be:

\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4

So then the answer is TRUE

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3 years ago
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\cfrac{x}{2}=\cfrac{\pi }{4}\implies \boxed{x=\cfrac{\pi }{2}}~\hfill \cfrac{x}{2}=\cfrac{3\pi }{4}\implies \boxed{x=\cfrac{3\pi }{2}}

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3 years ago
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<h3><u>Part A</u></h3>

\sf \rightarrow  y^2 - 3y - 4y +12

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collect into groups

\sf \rightarrow  (y- 4)(y - 3)

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collect into groups

\sf \rightarrow  (3x -2)( 3x-2)

simplify the following

\sf \rightarrow  (3x -2)^2

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2 years ago
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Contact [7]
<span>4k – 6 = -2k – 16 – 2
</span>4k+2k =  – 16 – 2+6
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5 0
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