Answer:
- <u><em>The solution to f(x) = s(x) is x = 2012. </em></u>
Explanation:
<u>Rewrite the table and the choices for better understanding:</u>
<em>Enrollment at a Technical School </em>
Year (x) First Year f(x) Second Year s(x)
2009 785 756
2010 740 785
2011 690 710
2012 732 732
2013 781 755
Which of the following statements is true based on the data in the table?
- The solution to f(x) = s(x) is x = 2012.
- The solution to f(x) = s(x) is x = 732.
- The solution to f(x) = s(x) is x = 2011.
- The solution to f(x) = s(x) is x = 710.
<h2>Solution</h2>
The question requires to find which of the options represents the solution to f(x) = s(x).
That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.
The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012,<em> x = 2012. </em>
Thus, the correct choice is the third one:
- The solution to f(x) = s(x) is x = 2012.
Answer:
3052. 08
Step-by-step explanation:
volume = 4/3πr^3
= 4/3π*(9)^3= 3052.08 in^3
If
is the amount of strontium-90 present in the area in year
, and it decays at a rate of 2.5% per year, then
![S(t+1)=(1-0.025)S(t)=0.975S(t)](https://tex.z-dn.net/?f=S%28t%2B1%29%3D%281-0.025%29S%28t%29%3D0.975S%28t%29)
Let
be the starting amount immediately after the nuclear reactor explodes. Then
![S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^{t+1}S(0)](https://tex.z-dn.net/?f=S%28t%2B1%29%3D0.975S%28t%29%3D0.975%5E2S%28t-1%29%3D0.975%5E3S%28t-2%29%3D%5Ccdots%3D0.975%5E%7Bt%2B1%7DS%280%29)
or simply
![S(t)=0.975^ts](https://tex.z-dn.net/?f=S%28t%29%3D0.975%5Ets)
So that after 50 years, the amount of strontium-90 that remains is approximately
![S(50)=0.975^{50}s\approx0.282s](https://tex.z-dn.net/?f=S%2850%29%3D0.975%5E%7B50%7Ds%5Capprox0.282s)
or about 28% of the original amount.
We can confirm this another way; recall the exponential decay formula,
![S(t)=se^{kt}](https://tex.z-dn.net/?f=S%28t%29%3Dse%5E%7Bkt%7D)
where
is measured in years. We're told that 2.5% of the starting amount
decays after 1 year, so that
![0.975s=se^k\implies k=\ln0.975](https://tex.z-dn.net/?f=0.975s%3Dse%5Ek%5Cimplies%20k%3D%5Cln0.975)
Then after 50 years, we have
![S(50)=se^{50k}\approx0.282s](https://tex.z-dn.net/?f=S%2850%29%3Dse%5E%7B50k%7D%5Capprox0.282s)
Answer:
about 18.2%
Step-by-step explanation:
All you have to do is divide the result of the percent, so 25.7, by the total, 141.
So 25.7÷141=0.18226950354
We round it to 0.182 and multiply it by 100 to know the percent which is 18.2