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Arlecino [84]
2 years ago
9

Time to paint the living room. The room measures 20 feet wide by 30 feet long by 12

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
6 0

The cost to paint the walls and the ceiling is $855

<h3></h3><h3>How much will it cost?</h3>

First, we need to find the area that will be painted.

The area of two of the walls is:

A = 20ft*12ft = 240ft²

The area of the other two walls:

A' = 30ft*12ft = 360ft²

The area of the ceiling is:

A'' = 20ft*30ft = 600ft²

We also need to subtract the areas of the two doors and the 4 windows.

Each door is 7ft by 3ft, so the area of each door is:

a = 7ft*3ft = 21ft²

Each window is 4ft by 3ft:

a' = 4ft*3ft = 12ft²

Then the total area that will be painted is:

area = 2A + 2A' + A - 2a - 4a'

area = 2*( 240ft²) + 2*(360ft²) + 600ft² - 2*(21ft²) - 4*(12ft²)

area = 1,710 ft²

We know that we need two coats of paint to cover that area, and the cost per square foot of paint is $0.25, then the total cost is:

C = $0.25*2*(1,710) = $855

The cost to paint the walls and the ceiling is $855

If you want to learn more about areas:

brainly.com/question/24487155

#SPJ1

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Attachment below

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The perimeter of a triangle is 17x−5 units. One side is 3x+5 units and another is 8x−3 units. How many units long is the third s
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17x−5 =85       3x+5 =15      8x-3= -24 here you can use this to find your awnser

Step-by-step explanation:

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Helppp me plsssssssss<br><br>​
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Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

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[0,12]

Step-by-step explanation:

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Answer:

Step-by-step explanation:

Okay!

A graph representing the function :

f(x)=x(x+2)

I will describe the graph for you!

the points on this graph are : (-3,3),(-2,0),(-1,-1),(0,0),(1,3)

the graph also creates a U shape!

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