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AVprozaik [17]
3 years ago
7

A cylindrical can of baked beans has a diameter of 8 cm and a height of 9 cm.

Mathematics
1 answer:
strojnjashka [21]3 years ago
7 0
Formula for volue of cylinder is area time height. Lets find area
A=\pi r^2
diameter=2r
8=2r
r=4
Now substitute r to A= \pi r^2
A=\pi 4^2= 16\pi
Now we have formula
V=16 \pi *height=16 \pi *9=144 \pi - its the result, so B 
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Step-by-step explanation:

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Write an inequality to describe the graph below ): tysm<33​
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y\leq \frac{3}{2}x+1

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Beverly is painting a mural on a 1-square-meter section of a wall. She has completed a rectangular part of the mural that's is 5
yawa3891 [41]

Answer:

0.21 square meter

Step-by-step explanation:

Area of the section of wall = 1 square meter

Dimension of part of mural completed = \frac{1}{4} meter by \frac{5}{6} meter

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But,

Area of a rectangle = length x width

So that,

Part of the mural completed = length x width

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The part of the mural that has been completed by Beverly is 0.21 square meter.

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3 years ago
The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room
Alexandra [31]

Answer:

\bar x = 260.1615

\sigma = 70.69

The confidence interval of standard deviation is: 53.76 to 103.25

Step-by-step explanation:

Given

n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

\bar x = \frac{\sum x}{n}

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

\bar x = \frac{5203.23}{20}

\bar x = 260.1615

\bar x = 260.16

Solving (b): The standard deviation

This is calculated as:

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

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}\sigma = \sqrt{\frac{94938.80}{19}}

\sigma = \sqrt{4996.78}

\sigma = 70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

So:

\alpha = 1 -c

\alpha = 1 -0.95

\alpha = 0.05

Calculate the degree of freedom (df)

df = n -1

df = 20 -1

df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

8 0
2 years ago
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