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2 years ago

14

Type the correct answer in the box. Jason builds doghouses for a pet store. Each doghouse is a wooden structure with a rectangul

ar base that has an area of 21 square feet and a length that is 4 feet more than its width. If x represents the width of the doghouse, write an equation in the given form that can be used to determine the possible dimensions of the base of the doghouse. 

2 answers:

Basile [38]2 years ago

8 0

Answer:

Width = 3 feet

Length = 7 feet

Step-by-step explanation:

x represents the width of doghouse so,

Width = x

Length is 4 times more than its width

Length = x+4

Area of doghouse = 21 square feet

We know

Area of rectangle = Length * Width

21 = (x+4)*x

21 = x^2+4x

=> x^2+4x-21=0

Solving the above quadratic equation by factorization to find the value of x

x^2+7x-3x-21=0

x(x+7)-3(x+7)=0

(x+7)(x-3)=0

x+7 =0 and x-3=0

x= -7 and x =3

Since the width of rectangle can never be negative so, x=3

Width =x = 3 feet

Length = x+4 = 3+4 = 7 feet

WINSTONCH [101]2 years ago

6 0

Answer: (x + 7)(x - 3) = 0

width (x) = 3

length (x+4) = 7

<u>Step-by-step explanation:</u>

Area (A) = length (l) × width (x)

21 = (x + 4) × (x)

21 = x² + 4x <em>distributed (x) into (x + 4)</em>

0 = x² + 4x - 21 <em>subtracted 21 from both sides</em>

0 = (x + 7)(x - 3) <em>factored quadratic equation</em>

0 = x + 7 or 0 = x - 3 <em>applied Zero Product Property</em>

x = -7 or x = 3 <em>solved each equation</em>

x = -7 is not valid <em>because lengths cannot be negative </em>

so x = 3

and length ... x + 4 = (3) + 4 = 7

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In which

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$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

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Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

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