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3 years ago

10

each pail of plaster covers 90 square feet of ceiling. What is the least number of pails of plaster you would need to buy to cov

er the ceiling of a room with walls 14 feet long?

2 answers:

Dominik [7]3 years ago

7 0

The area of the ceiling:
A = 14² = 196 ft²
We need plaster to cover 196 ft².
196 : 90 = 2.17778
Answer: We have to buy 3 pails of plaster.

noname [10]3 years ago

4 0

Answer:

3

Step-by-step explanation:

ik

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Answer:

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Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

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$Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}$

We need to find the lengths of side DE, EF and FD

Length of side DE = a = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side DE = a = $=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60$

Length of side EF = b = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side EF = b = $=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8$

Length of side FD = c = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side FD = c = $=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4$

so, a= 8.60, b= 5.8 and c = 4

s = a+b+c/2

s= 8.6+5.8+4/2

s= 9.2

Area of triangle= $=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88$

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Step-by-step explanation:

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Step-by-step explanation:

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Population mean, μ = 16.000

Sample mean, $\bar{x}$ = 16.218

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Alpha, α = 0.05

Sample standard deviation, s = 0.764

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