The answer is 95%
This is something you should memorize. Specifically it is from the Empirical Rule (or 68-95-99.7 rule) which gives rough approximations of areas under the curve.
Answer:
Width: 10 inches
Length: 15 inches
Step-by-step explanation:
Let the width be w.
Since the length is 5 inches longer than its width, the length = w +5.
Area of rectangle = length x width
w (w +5) = 150
w^2 + 5w = 150
w^2 + 5w -150 = 0
Using quadratic formula,
w = 10 or -15
Since the width cannot be negative,
the width is 10 inches.
Now just substitute w=10 into length = w +5.
length = 10 + 5
= 15 inches
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:
The mean of the numbers are 1
