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lord [1]
3 years ago
15

A painter needs to paint the bottom of a circular pool. The pool has a radius of 30 feet. A store sells 5-gallon cans of paint.

One gallon of paint covers 300 square feet. What is the smallest number of 5-gallon cans the painter must buy to cover the bottom of the pool? Explain or show your reasoning.
Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

2

Step-by-step explanation:

the area of the radius 30 is 2827 and since 1 bucket can cover 1800 we will only need 2 buckets because we need a total of 2827 and 1 bucket covers 1800 so 2 buckets will cover 3600.

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Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t
bogdanovich [222]

Answer:

a) \sigma_{\bar x} = 1.414

b) \sigma_{\bar x} = 1.414

c) \sigma_{\bar x} = 1.414

d) \sigma _{\bar x} = 1.343

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation  =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

c)  When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}

\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}

\sigma_{\bar x} = 1.414

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }

\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }

\sigma _{\bar x} = 1.343

7 0
3 years ago
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5 0
3 years ago
V7x – 1 = 3<br> Please help this is algebra 2
kumpel [21]

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6 0
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FrozenT [24]
<span>0.00006628369 is your answer for 0.45/6789</span>
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Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.
o-na [289]

Answer: \text{Length of AG=}\frac{2\sqrt{63}}{3}

Explanation:  

Please follow the diagram in attachment.  

As we know median from vertex C to hypotenuse is CM  

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Using trigonometry ratio identities  

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AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}

Length of AG=2/3 AN

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5 0
3 years ago
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