The probability is 1/1024.
Each tetrahedron has a 1/4 chance of landing on 3, since there are 4 sides and only one of them is marked 3.
Each tetrahedron roll is independent, since no roll is affected by another.
This means we multiply the probabilities:
1/4(1/4)(1/4)(1/4)(1/4) = 1/1024
$1250
Use the formula i = prt
In this case, we don't know P or Principal amount
i = 400
r = .04 *found by diving 4/100*
n = 8
Set formula to solve for P like: P= i/r(n)
P=400/.04(8)
P= 400/.32
P = 1250
To model half-life, use the formula
![A(t)=A_0 (\dfrac{1}{2})^{ \frac{t}{t_{1/2}}](https://tex.z-dn.net/?f=A%28t%29%3DA_0%20%28%5Cdfrac%7B1%7D%7B2%7D%29%5E%7B%20%5Cfrac%7Bt%7D%7Bt_%7B1%2F2%7D%7D%20)
. Here,
![A(t)](https://tex.z-dn.net/?f=A%28t%29)
is the amount remaining after a length of time
![t](https://tex.z-dn.net/?f=t)
.
![A_0](https://tex.z-dn.net/?f=A_0)
is the amount that you start with.
![t_{1/2}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D)
is the half-life. You plug in 50 for
![t_{1/2}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D)
, 10 for
![A_0](https://tex.z-dn.net/?f=A_0)
, and 25 for
![t](https://tex.z-dn.net/?f=t)
. You get
![A(t)=10( \frac{1}{2})^{25/50} = 10( \frac{1}{2})^{1/2} = 7.07 \ grams](https://tex.z-dn.net/?f=A%28t%29%3D10%28%20%5Cfrac%7B1%7D%7B2%7D%29%5E%7B25%2F50%7D%20%3D%2010%28%20%5Cfrac%7B1%7D%7B2%7D%29%5E%7B1%2F2%7D%20%3D%207.07%20%5C%20grams)
.
The answer is negative. You subtract 730 from 850
8r^6s^3 - 9r^5s^4 + 3r^4s^5 - (2r^4s^5 - 5r^3s^6 - 4r^5s^4) =
8r^6s^3 - 9r^5s^4 + 3r^4s^5 - 2r^4s^5 + 5r^3s^6 + 4r^5s^4 =
8r^6s^3 -5r^5s^4 + r^4s^5 + 5r^3s^6 <==