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Marat540 [252]
1 year ago
10

If w represents the width of the tree house, which inequality could be used to determine what lengths would make the area of the

base of the tree house greater than 293 square inches?
Mathematics
1 answer:
Doss [256]1 year ago
7 0

An inequality which could be used to determine the lengths that would make the area of the rectangular base of the tree house greater than 293 square inches is W² + 7W > 293.

<h3>How to calculate the area of a rectangle?</h3>

Mathematically, the area of a rectangle can be calculated by using this formula;

A = LW

Where:

  • A is the area of a rectangle.
  • L is the length of a rectangle.
  • W is the width of a rectangle.

For the length and area of the rectangular base, we have:

L = W + 7

A > 293

Substituting the given parameters into the formula, we have;

L × W > 293

(W + 7) × W > 293

W² + 7W > 293

Read more on area of a rectangle here: brainly.com/question/25292087

#SPJ4

Complete Question:

Adam built a tree house with a rectangular base. The length of the base is 7 inches more than its width. If w represents the width of the tree house, which inequality could be used to determine what lengths would make the area of the base of the tree house greater than 293 square inches?

W + 7 > 293

202 + 286 > 2,051

w² + 7w > 293

12 + 293 > 293

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andrey2020 [161]

Answer:

(16 x^8 - x^3 + 6)/(2 x^3)

Step-by-step explanation:

Simplify the following:

(64 x^8 - 4 x^3 + 24)/(8 x^3)

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(4 (16 x^8 - x^3 + 6))/(8 x^3)

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Answer: (16 x^8 - x^3 + 6)/(2 x^3)

7 0
3 years ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

Yes the students are reasonably good at estimating one​ minute

a

  Option A is  correct

b

  The test statistics is  t  =  0.354

Step-by-step explanation:

From the question we are told that

    The  set of data is  

               68, 82 ,  38 ,  62 , 41, 25 , 57 ,  64, 67, 47, 61, 71, 91, 87, 64

     The  population mean is  \mu  = 60

The  level of significance is given as\alpha =  0.01

    The  critical value for this level of significance obtained from the normal distribution table is  

         Z_{\alpha } =  2.33

   The  null hypothesis is  

           H_o  :  \mu  =  60 \ seconds

   The alternative hypothesis is  

          Ha :  \mu \ne 60 \ seconds      

Generally the sample mean is mathematically represented as

       \= x  =  \frac{\sum x_i }{n  }

where  n = 15

 So  

      \= x  =  \frac{ 68+ 82 +  38 +  62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}

      \= x  =  61.67

The standard deviation is mathematically represented as

     \sigma  =\sqrt{  \frac{ \sum  (x_i -  \= x )^2}{n} }

substituting values

      \sigma  =\sqrt{  \frac{ \sum  (68-61.67)^2 + ( 82-61.67 )^2  +   (38 -61.67)^2 +  62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }           \sqrt{  \frac{  ( \cdot  \cdot  + (  64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+  (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 +  (64-61.67)^2}{15} }=>   \sigma  =  18.23

The  test statistics is evaluated as  

      t  =  \frac{\= x  - \mu  }{ \frac{\sigma }{\sqrt{n} } }

substituting values

       t  =  \frac{61.67  -60  }{ \frac{18.23 }{\sqrt{ 15} } }

      t  =  0.354

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

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Answer:

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Answer:

Step-by-step explanation:

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Answer:

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