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77julia77 [94]
3 years ago
7

What is the volume of a cylinder with a radius of 7 and the height of 9​

Mathematics
2 answers:
castortr0y [4]3 years ago
5 0

Answer: 1,384.74 units³

Step-by-step explanation: To find the volume of a cylinder, start with the formula for the volume of a cylinder.

Volume = πr²h

Notice that our cylinder has a radius of 7 units and a height of 9 units so plugging into the formula, we have (π)(7 units)²(9 units).

Start by simplifying the exponent. 7² is equal to 7 units × 7 units or 49 units² so we have (π)(49)(9). Now, 49 × 9 is 441π units³.

So the volume of the cylinder is 441π units³.

Now, remember that π is approximately equal to 22/7 or 3.14 so we can estimate the value of the volume by plugging in 3.14 for π.

So we have (441)(3.14) which is equal to 1,384.74 units³.

So the value of the volume is approximately 1,384.74 units³.

xenn [34]3 years ago
4 0

volume of a cylinder: πr²h

volume = π(7²)(9) = 441π = 1385.44 un² approx.

answer: 1385.44 un² approx.

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Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

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The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

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d) (1) 0,10,50,60,100

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Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

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