Answer:
use formula length × width ×height
Step-by-step explanation:
plug in what they have given you
Answer:
14
Step-by-step explanation:
9 + 4 = 13 and 1/2 + 1/2 = 2/2 = 1 and 13 + 1 = 14
Check the picture below. When will it reach its height? well, at the vertex, so
![~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&24\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&50\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+24t+50\implies \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%2624%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%2650%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20h%28t%29%3D-16t%5E2%2B24t%2B50%5Cimplies%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D)
![h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+24}t\stackrel{\stackrel{c}{\downarrow }}{+50} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 24}{2(-16)}~~~~ ,~~~~ 50-\cfrac{ (24)^2}{4(-16)}\right)\implies \left( \cfrac{3}{4}~~,~~50-\cfrac{576}{64} \right)](https://tex.z-dn.net/?f=h%28t%29%3D%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B-16%7Dt%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B24%7Dt%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B%2B50%7D%20%5Cqquad%20%5Cqquad%20%5Cleft%28-%5Ccfrac%7B%20b%7D%7B2%20a%7D~~~~%20%2C~~~~%20c-%5Ccfrac%7B%20b%5E2%7D%7B4%20a%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28-%5Ccfrac%7B%2024%7D%7B2%28-16%29%7D~~~~%20%2C~~~~%2050-%5Ccfrac%7B%20%2824%29%5E2%7D%7B4%28-16%29%7D%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B3%7D%7B4%7D~~%2C~~50-%5Ccfrac%7B576%7D%7B64%7D%20%5Cright%29)
![\left(\cfrac{3}{4}~~,~~50+9 \right)\implies \stackrel{\underset{\qquad \downarrow }{\textit{how far up it went}}}{\underset{\stackrel{\uparrow \qquad ~~}{\textit{when it reached it}}}{\left( \cfrac{3}{4}~~,~~59 \right)}}](https://tex.z-dn.net/?f=%5Cleft%28%5Ccfrac%7B3%7D%7B4%7D~~%2C~~50%2B9%20%5Cright%29%5Cimplies%20%5Cstackrel%7B%5Cunderset%7B%5Cqquad%20%5Cdownarrow%20%7D%7B%5Ctextit%7Bhow%20far%20up%20it%20went%7D%7D%7D%7B%5Cunderset%7B%5Cstackrel%7B%5Cuparrow%20%5Cqquad%20~~%7D%7B%5Ctextit%7Bwhen%20it%20reached%20it%7D%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B3%7D%7B4%7D~~%2C~~59%20%5Cright%29%7D%7D)
Answer:
a) (490,510)
b) (480,520)
c) (470,530)
Step-by-step explanation:
Mean, μ = 500
Standard Deviation, σ = 10
Empirical Rule:
- According to Empirical rule almost all data lies within three standard deviation of mean for a normal distribution.
- About 68% of data lies within one standard deviation of the mean.
- About 95% of data lies within two standard deviation of mean.
- About 99.7% of data lies within three standard deviation of mean.
a) 68% of the observations
![\mu \pm \sigma\\=500 \pm 10\\=(490, 510)](https://tex.z-dn.net/?f=%5Cmu%20%5Cpm%20%5Csigma%5C%5C%3D500%20%5Cpm%2010%5C%5C%3D%28490%2C%20510%29)
About 68% of data lies within 490 and 510.
b) 95% of the observations
![\mu \pm 2\sigma\\=500 \pm 2(10)\\=(480, 520)](https://tex.z-dn.net/?f=%5Cmu%20%5Cpm%202%5Csigma%5C%5C%3D500%20%5Cpm%202%2810%29%5C%5C%3D%28480%2C%20520%29)
About 95% of data lies within 480 and 520.
c) All the observations
![\mu \pm 3\sigma\\=500 \pm 3(10)\\=(470, 530)](https://tex.z-dn.net/?f=%5Cmu%20%5Cpm%203%5Csigma%5C%5C%3D500%20%5Cpm%203%2810%29%5C%5C%3D%28470%2C%20530%29)
All of the observations lie between 470 and 530.