Answer:
63,62,7,192
Step-by-step explanation:
1.63
2.62
3.7
4.192 (some of it got cut off : ) )
bearing in mind that 4¾ is simply 4.75.
![\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%5C%24600%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20n%3D%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%20%5Ctextit%7Bannually%2C%20thus%20once%7D%20%5Cend%7Barray%7D%5Cdotfill%20%261%5C%5C%20t%3Dyears%5Cdotfill%20%263%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D600%5Cleft%281%2B%5Cfrac%7B0.05%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%203%7D%5Cimplies%20A%3D600%281.05%29%5E3%5Cimplies%20A%3D694.575%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

well, the interest for each is simply A - P
695.575 - 600 = 95.575.
862.032 - 750 = 112.032.
Answer:
85 flu patients Per week so A
Step-by-step explanation:
Answer:
160
Step-by-step explanation:
10% of the candies is 16, and to get 100% from 10% is to multiply by 10.
So 10% *10 = 100%, then by association if 16 = 10%, then 16*10 will be the total 100%.
Well first find the slope of the point
y=mx+b y=-1x+3
slope = -1
then put the problem in point slope form y - y1 = m (x-x1)
y + 3 = -1 (x+1) then solve for y=mx+b again
so y> 2x + 6 is the answer