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Serhud [2]
3 years ago
12

Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p

ut in the pen which has a perimeter that is equal to the area of its base. After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen. He also finds that the perimeter of the pen is 3 times greater than the perimeter of the doghouse. If x represents the length of the pen and y represents the area of the doghouse, then which of the following systems of equations can be used to determine the length and width of the pen and the area of the doghouse?
Options:

A)
178=x²-3x-y
6=4x-3y

B)
178=x²+3x-y
-6=4x-3y

C)
178=x²+3x
-6=4x-3y

D)
178=x²+3x+y
-6=4x-3y

Mathematics
2 answers:
zalisa [80]3 years ago
8 0

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

<u>Statement 1: </u>"The pen is 3 feet wider than it is long"

Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3

------

<u>Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"</u>

Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y

------

<u>Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."</u>

Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\

------

<u>Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."</u>

Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\

Conclusion:

The systems of equations that can be used to determine the length and width of the pen and the area of the doghouse is given in Option B.

178=x^2+3x-y\\ \\ -6=4x-3y

THCbbb2 years ago
0 0

thank you very muvh

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c)

According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

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Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

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3  0.001087  

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7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

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