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larisa [96]
3 years ago
12

A room is 18 ft long, 14 ft wide, and 8 ft high.

Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0
A) The answer is $38.95
Two walls have dimensions: 14 ft x 8 ft, each with area A1 = 14 · 8 = 112 ft²
Two walls have dimensions: 18 ft x 8 ft each with area A2 = 18 · 8 = 144 ft²
The area that needs to be painted is:
A = 2A1 + 2A2 = 2*112 + 2*144 = 224 + 288 = 512 ft²
Since you need two coats of paint, total area that needs to be painted is
512 × 2 = 1024 ft²

Now, make a proportion: 
If 1 gallon covers 250 ft², how many gallons will cover 1024 ft²
1 : 250 = x : 1024
x = 1024/250 = 4.096 gallons ≈ 4,1 gallons

Since 1 gallon costs $9.50, 4.1 gallons will cost $38.95:
$9.50 × 4.1 gallons = $38.95


b) The answer is $1,260.
The dimensions of the floor are: 18 ft × 14 ft
Therefore, the area that needs to be covered with carpet is:
A = 18 · 14 = 252 ft²
Since 1 ft² costs $5, that means that 252 ft² will cost $1,260:
252 ft² × $5 = $1,260


c) The answer is $1,890.
The ceiling will have the same dimensions and the same area as the floor.
The dimensions of the ceiling are: 18 ft × 14 ft
Therefore, the area that needs to be covered is:
A = 18 · 14 = 252 ft²
Since 1 ft² costs $7.50, that means that 252 ft² will cost $1,890:
252 ft² × $7.5 = $1,890


d) The answer is $3,188.95.
We now know that:
- the cost of painting the four walls with two coats of paint is: $38.95
- the cost of carpeting the floor with carpet is: $1,260
- the cost of ceiling with acoustic tile is: $1,890
All we have to do is to sum it up to find the total cost of renovating the walls,floor, and ceiling:
$38.95 + $1,260 + $1,890 = $3,188.95
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80 is 32% of what number
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The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab
Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

\pi = 0.25

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when M = 0.1

So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}

0.1\sqrt{n} = 2.575{0.25*0.75}

\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}

(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}

n = 124.32

Rounding up

The minimum sample size needed is 125.

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Its D(5,-4) XD ;p
because I did it and got it right but don't take my word for it just ask the other suggestions <span />
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