so, we have two 54x18 rectangles, so their perimeter is simply all those units added together, 54+54+54+54+18+18+18+18 = 288.
we know the circle's diameter is 1.5 times the width, well, the width is 18, so the diameter of the circle must be 1.5*18 = 27.
![\bf \stackrel{\textit{circumference of a circle}}{C=d\pi }~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=27 \end{cases}\implies C=27\pi \implies C=\stackrel{\pi =3.14}{84.78} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perimeter of the rectangles}}{288}~~~~+~~~~\stackrel{\textit{perimeter of the circle}}{84.78}~~~~=~~~~372.78](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bcircumference%20of%20a%20circle%7D%7D%7BC%3Dd%5Cpi%20%7D~~%20%5Cbegin%7Bcases%7D%20d%3Ddiameter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20d%3D27%20%5Cend%7Bcases%7D%5Cimplies%20C%3D27%5Cpi%20%5Cimplies%20C%3D%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B84.78%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20rectangles%7D%7D%7B288%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20circle%7D%7D%7B84.78%7D~~~~%3D~~~~372.78)
Answer:
look it up
Step-by-step explanation:
L.O.O.K I.T U.P.
Answer:
0.683
Step-by-step explanation:
We have to find P(-1<z<1).
For this purpose, we use normal distribution area table
P(-1<z<1)=P(-1<z<0)+(0<z<1)
Using normal area table and looking the value corresponds to 1.0, we get
P(-1<z<1)=0.3413+0.3413
P(-1<z<1)=0.6826
Rounding the answer to three decimal places
P(-1<z<1)=0.683
So, 68.3% of the z-scores will be between -1 an 1.
69 babyyyyyyyyyyyyyyyyyyyyyy
Answer:
x=8, y=-1
Step-by-step explanation:
The figure is said to be a rectangle, which means opposite sides have to be equal in length. From this, we can write a system of equations:
5x-2y=42
2x+3y=13
There are a number of ways to solve this but I'll go with elimination--that means you need to try to remove one variable first by manipulating the equations and solve for the other variable. Then you use the solution to solve for the variable you initially removed. That sounds confusing, but let me show you how to do it.
If you multiply the first equation by 3 and the second equation by 2, you'll have have:
15x-6y=126
4x+6y=26
Now, we can add these two equations. The y will cancel out, so we can easily solve for x.
19x=152
x=8
Use this value of x to substitute back into either of the original equations to find y.
5(8)-2y=42
40-2y=42
-2y=2
y=-1
