Check the picture below.
so, let's notice, is really just a 2x20 rectangle with a quarter of a semicircle with a radius of 11.
![\bf \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \implies A=\pi 11^2\implies A=121\pi \implies \stackrel{\textit{one quarter of that}}{\boxed{A=\cfrac{121\pi }{4}}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\underline{\textit{area of the figure}}}{\stackrel{\textit{rectangle's area}}{(2\cdot 20)}+\stackrel{\textit{circle's quart's area}}{\cfrac{121\pi }{4}}\qquad \approx \qquad 135.03\implies \stackrel{\textit{rounded up}}{135}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cimplies%20A%3D%5Cpi%2011%5E2%5Cimplies%20A%3D121%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7B%5Cboxed%7BA%3D%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Cunderline%7B%5Ctextit%7Barea%20of%20the%20figure%7D%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Brectangle%27s%20area%7D%7D%7B%282%5Ccdot%2020%29%7D%2B%5Cstackrel%7B%5Ctextit%7Bcircle%27s%20quart%27s%20area%7D%7D%7B%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%5Cqquad%20%5Capprox%20%5Cqquad%20135.03%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B135%7D%7D)
Answer:
f(x)= (-4)/(5-X)
Step-by-step explanation:
Answer:
3 times 4 over 5 equals 2.4
Step-by-step explanation:
this Is a semi easy one the answer is 21. here is how you do it.
first change 56% into a decimal which comes out to be .56 now subtract it from 100% but in decimal form which is .1.
.1-.56=.44
now you want to take that .44 and multiply it by 14 which comes out to be 6.16 but you cant have .16th of a person so lets just say its 7 people.
now add 7 to the 14 clowns and you get 21.
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The interval would be from 87.8 to 124.8.
68% of the data fall within one standard deviation of the mean, so we calculate
106.3 +/- 18.5:
106.3 - 18.5 = 87.8
106.3 + 18.5 = 124.8