It would make you less likely because you might have to pay the money you were gonna pay to the people who did the car loan.
Answer:
umm what or we doing?
Step-by-step explanation:
This is a scholarship.
Student loans must be repaid; grants are given usually without those types of restrictions; and student work involves a student having an on-campus job.
To solve this we are going to use the half life equation
![N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }](https://tex.z-dn.net/?f=N%28t%29%3DN_%7B0%7D%20e%5E%7B%28%20%5Cfrac%7B-0.693t%7D%7Bt%20_%7B1%2F2%7D%20%7D%29%20%7D%20)
Where:
![N_{0}](https://tex.z-dn.net/?f=N_%7B0%7D%20)
is the initial sample
![t](https://tex.z-dn.net/?f=t)
is the time in years
![t_{1/2}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20)
is the half life of the substance
![N(t)](https://tex.z-dn.net/?f=N%28t%29)
is the remainder quantity after
![t](https://tex.z-dn.net/?f=t)
years
From the problem we know that:
![N_{0} =100](https://tex.z-dn.net/?f=N_%7B0%7D%20%3D100)
![t=200](https://tex.z-dn.net/?f=t%3D200)
![t_{1/2} =1600](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D1600)
Lets replace those values in our equation to find
![N(t)](https://tex.z-dn.net/?f=N%28t%29)
:
![N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }](https://tex.z-dn.net/?f=N%28200%29%20%3D100e%5E%7B%28%20%5Cfrac%7B%28-0.693%29%28200%29%7D%7B1600%7D%29%20%7D%20)
![N(200)=100e^{( \frac{-138.6}{1600} )}](https://tex.z-dn.net/?f=N%28200%29%3D100e%5E%7B%28%20%5Cfrac%7B-138.6%7D%7B1600%7D%20%29%7D%20)
![N(200)=100e^{-0.086625}](https://tex.z-dn.net/?f=N%28200%29%3D100e%5E%7B-0.086625%7D%20)
![N(200)=91.7](https://tex.z-dn.net/?f=N%28200%29%3D91.7)
We can conclude that after 1600 years of radioactive decay, the mass of the 100-gram sample will be
91.7 grams.
Answer:Im pretty sure it's 25% percent loss
Step-by-step explanation:
sorry if im wrong