lemme use a slightly different equation, just the variables differ, but is basically the same you have there.
A)
![\bf \textit{Amount of Population Growth, \boxed{\textit{10th day}}} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&5287\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &10\\ \end{cases} \\\\\\ A=5287e^{10r} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BAmount%20of%20Population%20Growth%2C%20%5Cboxed%7B%5Ctextit%7B10th%20day%7D%7D%7D%20%5C%5C%5C%5C%20A%3DPe%5E%7Brt%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%265287%5C%5C%20P%3D%5Ctextit%7Binitial%20amount%7D%5Cdotfill%20%26P%5C%5C%20r%3Drate%5Cto%20r%5C%25%5Cto%20%5Cfrac%7Br%7D%7B100%7D%5C%5C%20t%3D%5Ctextit%7Belapsed%20time%7D%5Cdotfill%20%2610%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D5287e%5E%7B10r%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \textit{Amount of Population Growth, \boxed{\textit{21st day}}} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&692\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &21\\ \end{cases} \\\\\\ 692=Pe^{21r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 5287=Pe^{10r}\\ 692=Pe^{21r} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BAmount%20of%20Population%20Growth%2C%20%5Cboxed%7B%5Ctextit%7B21st%20day%7D%7D%7D%20%5C%5C%5C%5C%20A%3DPe%5E%7Brt%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%26692%5C%5C%20P%3D%5Ctextit%7Binitial%20amount%7D%5Cdotfill%20%26P%5C%5C%20r%3Drate%5Cto%20r%5C%25%5Cto%20%5Cfrac%7Br%7D%7B100%7D%5C%5C%20t%3D%5Ctextit%7Belapsed%20time%7D%5Cdotfill%20%2621%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20692%3DPe%5E%7B21r%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%205287%3DPe%5E%7B10r%7D%5C%5C%20692%3DPe%5E%7B21r%7D%20%5Cend%7Bcases%7D)
![\bf ln\left( \cfrac{692}{5287}\right)=ln\left( e^{11r} \right)\implies ln\left( \cfrac{692}{5287}\right)=11r\implies \cfrac{ln\left( \frac{692}{5287}\right)}{11}=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -0.18486\approx r~\hfill](https://tex.z-dn.net/?f=%5Cbf%20ln%5Cleft%28%20%5Ccfrac%7B692%7D%7B5287%7D%5Cright%29%3Dln%5Cleft%28%20e%5E%7B11r%7D%20%5Cright%29%5Cimplies%20ln%5Cleft%28%20%5Ccfrac%7B692%7D%7B5287%7D%5Cright%29%3D11r%5Cimplies%20%5Ccfrac%7Bln%5Cleft%28%20%5Cfrac%7B692%7D%7B5287%7D%5Cright%29%7D%7B11%7D%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20-0.18486%5Capprox%20r~%5Chfill)
B)
and we can round that up to a whole leaf of 6361, or truncate it to 6360, chances are is the latter.
C)
and we can round that up to 47 days even.
Answer:
The mean will change. The range will change.
Step-by-step explanation:
The outlier is 4, it's the one number in the data that isn't close to the other numbers.
The data without the outlier will be 7, 9, 9, 10.
Mode: The mode will still be 9. It is still the number that appears the most in the data.
Median: The median WON'T change. The data is now one value shorter, but the median will still be 9. Before it was 9, and now it's still 9 ((9+9)/2).
Mean: The mean will change because we're excluding one value. Before, the mean was 7.8 (4+7+9+9+10=39, 39/5 = 7.8). NOW it is 8.75 (7+9+9+10=35, 35 / 4 = 8.75).
Range: The range will change. Before it was 10-4, now it is 10-7. The range was originally 6 and now it's 3.
He liked about 8-9 colors because all you have to do is subtract 33% which equals 8.71
Answer:
A U B= 1,2,3,4,6,10,11
Step-by-step explanation:
because A union B is all number from set A and set B together without repeating
Answer:
Step-by-step explanation:
Hello,
let s assume that m+n is different from 0
we have this equation and we need to find m as a function of t, s, and n
<=>
for t-s different from 0, so t different from s
hope this helps
