C. ED NEEDS ME TO MAKE IT LONGER
To find the surface area of a cylinder add the surface area of each end plus the surface area of the side. Each end is a circle so the surface area of each end is π * r2, where r is the radius of the end. There are two ends so their combinded surface area is 2 π * r2. The surface area of the side is the circumference times the height or 2 π * r * h, where r is the radius and h is the height of the side.
The entire formula for the surface area of a cylinder is 2 π r2 + 2 π r h
Answer:
-1.25 :)
Step-by-step explanation:
A<span>. What proportion have SAT scores greater than 700? Z= 2 ; 2.28%
B. What proportion have SAT scores greater than 500? 50%
C. What is the minimum SAT score needed to be in the highest
10% of the population? Z= 1.28; 628
D. If the state college only accepts students from the top 60% of the SAT distribution, what is the minimum SAT score needed to be accepted? Z= .26; 474</span><span>
</span>
Check the picture below.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}asn ~~ \begin{cases} a=apothem\\ n=\stackrel{side's}{number}\\ s=\stackrel{side's}{length}\\[-0.5em] \hrulefill\\ a=7\\ s=8.1\\ n=6 \end{cases}\implies \stackrel{\textit{area of the hexagonal base}}{A=\cfrac{1}{2}(7)(8.1)(6)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dasn%20~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20n%3D%5Cstackrel%7Bside%27s%7D%7Bnumber%7D%5C%5C%20s%3D%5Cstackrel%7Bside%27s%7D%7Blength%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D7%5C%5C%20s%3D8.1%5C%5C%20n%3D6%20%5Cend%7Bcases%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagonal%20base%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{volume of a prism}\\\\ V=Bh ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\frac{1}{2}(7)(8.1)(6)\\ V=3572.1 \end{cases}\implies 3572.1=\cfrac{1}{2}(7)(8.1)(6)h \\\\\\ 3572.1=170.1h\implies \cfrac{3572.1}{170.1}=h\implies \boxed{21=h}](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20prism%7D%5C%5C%5C%5C%20V%3DBh%20~~%20%5Cbegin%7Bcases%7D%20B%3D%5Cstackrel%7Bbase%27s%7D%7Barea%7D%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20B%3D%5Cfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29%5C%5C%20V%3D3572.1%20%5Cend%7Bcases%7D%5Cimplies%203572.1%3D%5Ccfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29h%20%5C%5C%5C%5C%5C%5C%203572.1%3D170.1h%5Cimplies%20%5Ccfrac%7B3572.1%7D%7B170.1%7D%3Dh%5Cimplies%20%5Cboxed%7B21%3Dh%7D)
well, the length of the tunnel is "h", now two 8 meters cars, that's 8+8=16 meters plus a 3 meter connector between them, that's 16 + 3 = 19 meters, can those two cars connected like so fit inside the tunnel? sure thing, "h" can fit 19 meters just fine.
