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MrRa [10]
2 years ago
11

Find the volume of the cone

Mathematics
1 answer:
NeTakaya2 years ago
8 0

The volume of the cone is \frac{1}{3} \pi r^{2}h

  • r: radius --> 6ft
  • h: height --> 6ft
  • \pi = 3.14

Volume = \frac{1}{3} \pi (6)^{2} *6=226.08ft^{3}

Hope that helps!

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A rectangular prism must have a base with an area of no more than 27 square meters. The width of the base must be 9 meters less
den301095 [7]

Answer:

The maximum height of the prism is 12\ m

Step-by-step explanation:

Let

x------> the height of the prism

we know that

the area of the rectangular base of the prism is equal to

A=L*W

A\leq 27\ m^{2}

so

L*W\leq 27 -------> inequality A

W=x-9 ------> equation B

L=W+6 -----> equation C

Substitute equation B in equation C

L=(x-9)+6

L=x-3 ------> equation D

Substitute equation B and equation D in the inequality A

(x-3)*(x-9)\leq 27-------> using a graphing tool to solve the inequality

The solution for x is the interval---------->[0,12]

see the attached figure

but remember that

The width of the base must be 9 meters less than the height of the prism

so

the solution for x is the interval ------> (9,12]

The maximum height of the prism is 12\ m

8 1
2 years ago
Read 2 more answers
Rewrite 1/2 3/4 as a unit rate ​
Nata [24]

Answer:

Step-by-step explanation yes

5 0
3 years ago
You and a friend are playing a best-of-three series (first to win 2 games wins the series). You are a stronger player and have a
mariarad [96]

Answer and Step-by-step explanation:

Given that probability of you winning each game = 0.68

And probability of you winning next game = 0.81

Your friend's chance of winning/you losing would be = 1-0.68= 0.32

also his chance of winning next game = 0.73

To find probability that you would win the series given that you need to win two games to win the series

= probability that you win first game and second game+ probability that you win first game, lose second game and win third game + probability that you lose first game, win second game and win third game

= 0.68*0.81+0.68*0.32*0.68+0.32*0.68*0.81

=0.8750

Therefore probability that you would win the series = 0.8750

Note: here we found the probability of winning by adding(or) up the three possible combinations that would result in a win

6 0
2 years ago
For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;
umka2103 [35]

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

IQR = 13.5 - 4 = 9.5

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

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

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) (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) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

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

IQR = 8.5 - 4 = 4.5

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

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

Median = 3rd variable = 50

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.

Hope this Helps!!!

5 0
3 years ago
At a charity fundraiser, each female attendant donated $1,200 and each male attendant donated $800. If the average donation at t
Flauer [41]

The ratio of the number of female attendants to the number of male attendants is 21/17

<h3>Ratio</h3>

  • Amount donated by female attendant = $1200
  • Amount donated by male attendant = $800
  • Average donation = $900

let

  • number of female attendant = x
  • number of male attendant = y

900 = 1200x + 800y / (x + y)

900x + 900y = 1200x + 800y

900x + 1200x = 800y + 900y

2100x = 1700y

x : y = 2100 / 1700

x : y = 21 ; 17

Learn more about ratio:

brainly.com/question/2784798

#SPJ1

3 0
2 years ago
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