let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.
think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.
now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D12%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%2012%5Cimplies%20C%3D24%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthree%20quarters%20of%20it%7D%7D%7B24%5Cpi%20%5Ccdot%20%5Ccfrac%7B3%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%206%5Cpi%20%5Ccdot%203%5Cimplies%2018%5Cpi)
well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20C%3D18%5Cpi%20%5Cend%7Bcases%7D%5Cimplies%2018%5Cpi%20%3D2%5Cpi%20r%5Cimplies%20%5Ccfrac%7B18%5Cpi%20%7D%7B2%5Cpi%20%7D%3Dr%5Cimplies%209%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bdiameter%20is%20twice%20the%20radius%7D%7D%7Bd%3D18%7D~%5Chfill)
Answer:
- <u>The correct statement is the first one: </u><u><em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean</em></u><em> </em>
<em />
Explanation:
To calculate how many<em> standard deviations</em> a particular value in a group is from the mean, you can use the z-score:
Where:
is the number of standard deviations the value of x is from the mean
is the mean
is the standard deviation
Substitute in the formula:
Which means that <em>the number of blue-eyed students in Mr. Garcia's class is 2 standard deviations</em> above the mean.
Above the mean is the same that to the right of the mean, because the in the normal standard probability graph the central value is Z = 0 (the z-score of the mean value is 0), the positive values are to the right of the central value, and the negative values are to the left of the central value.
Therefore, the correct statement is the first one: <em>The number of blue-eyed students in Mr. Garcia's class is 2 standard deviations to the right of the mean, </em>
Answer:I think that it would be either b or c
Step-by-step explanation:
Assuming I understand your question correctly, in that you’re looking for just some descriptions of the differences between the functions. If so, then I’d say:
First graph both functions, the f(x) and the g(x). Then spot the differences.
Note that the g(x) function has shifted towards the right compared with the f(x) function.
Another way that the g(x) differs from the f(x) function is that it’s stretched. The vertex is in the IV quadrant for g(x) rather than at the origin for f(x).
I hope that helps.
Answer:
Plain is $7 and Holiday is $15
Step-by-step explanation:
Castel: 2p + 5h = 89
Kali: 9p + 10h = 213
Double Castel's: 4p + 10h = 178
Subtract fro Kali's 9p + 10h = 213
- 4p + 10h = 178
5p = 35
p = 7 Plain is $7
Substitute 7 into one of the equations and solve for h:
2(7) + 5h = 89
14 + 5h = 89
5h = 75
h = 15 Holiday is $15
