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:
18 minutes
Step-by-step explanation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is the time,
and T is the half life.
A = 140 when t = 10. Solve for the half life:
140 = 700 (½)^(10 / T)
0.2 = ½^(10 / T)
log 0.2 = (10 / T) log 0.5
10 / T = 2.32
T = 4.31
When A = 40, t is:
40 = 700 (½)^(t / 4.31)
0.057 = ½^(t / 4.31)
log 0.057 = (t / 4.31) log 0.5
t / 4.31 = 4.13
t = 17.8
Rounded to the nearest whole number, it takes 18 minutes.
Answer:
.6, 6/10, 15/25
Step-by-step explanation:
hope this helps!❄︎