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kompoz [17]
3 years ago
6

Alice purchased paint in a bucket with a radius of 9 in. and a height of 9.5 in. The paint cost $0.09 per in3.

Mathematics
1 answer:
coldgirl [10]3 years ago
5 0
\pir^2 = cross-sectional area of paintbucket.
3.14 (Your approximate value of pi) * 9^2 = 254.34
254.34 * 9.5 = total volume of bucket.
2,416.23 in^3.

0.09$ * 2416.23 = cost
The cost would be $217.46 (To the nearest cent)
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notsponge [240]
The answer is D
you would set up an equation like this
14,250 x .032
this will give you how much the population is decreasing per year. Which is 456 people per year.
You then take 2015 and find the difference between that number and 2025 and do it with all of the other years too.
For example 2025-2015 is 10 years
You then take 10 x 456 to tell you how much the of population you lost in those 10 years. You will then take that number which is 4,560 and do 14,250-4,560 to get 9,650.
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When did my mom chop up my glizzy<br><br> A:2015<br> B:2017<br> C:2018<br> D:2020
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Answer:

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5 0
3 years ago
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Solve A = a+b/2 for b
jeka94

Answer:

Step-by-step explanation:

I'm not sure if the right side is \frac{a + b}{2} or a + \frac{b}{2}, so I'll provide an answer for both:

A = \frac{a + b}{2}

2A = a + b

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6 0
3 years ago
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Below are the jersey numbers of 11 players randomly selected from a football team. Find the​ range, variance, and standard devia
jeyben [28]

Answer:

The range is 77.

The variance is 739.8.

The standard deviation is 27.2.

Step-by-step explanation:

The data provided, in ascending order is as follows:

S = {11 , 14 , 23 , 31 , 40 , 60 , 63 , 68 , 75 , 77 , 88}

The range of a data set is the difference between the maximum and minimum values.

From the above data set we know,

Maximum = 88

Minimum = 11

Compute the range as follows:

Range = Maximum - Minimum

           = 88 - 11

           = 77

The range is 77.

Compute the mean of the data as follows:

\bar x=\frac{1}{n}\sum x_{i}=\frac{1}{11}\times [11+14+23+...+88]=50

Compute the variance as follows:

\sigma^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x})^{2}

    =\frac{1}{11-1}\times [(11-50)^{2}+(14-50)^{2}+...+(88-50)^{2}]\\\\=\frac{1}{10}\times 7398\\\\=739.8

The variance is 739.8.

Compute the standard deviation as follows:

\sigma=\sqrt{\frac{1}{n-1}\sum (x_{i}-\bar x})^{2}}=\sqrt{739.8}=27.199265\approx 27.2

The standard deviation is 27.2.

4 0
3 years ago
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