Order of operations (PEMDAS):
21 - 18 <span>/ 3 * 2
let's add some parenthesese to make it easier to see:
21 - (18 / 3 * 2)
multiplication and division are associative
also: 18 / 3 = 18 * (1/3)
21 - (18 * (1/3) * 2)
21 - (6*2) or 21 - (36 * (1/3)) or 21 - (18 * (2/3))
21 - 12
9
</span>
I basically converted them all into decimals and rounded
Team 1- 0.75
Team 2- 0.8
Team 3- 0.775
Team 4- 0.825
Team 5- 0.725
Team 6- 0.7625
Team 7- 0.7875
Team 8- 0.8125
Team 4 collected the most
Team 5 collected the least
5.45,5.451,234,579..............
Answer:
Q(t) = Q_o*e^(-0.000120968*t)
Step-by-step explanation:
Given:
- The ODE of the life of Carbon-14:
Q' = -r*Q
- The initial conditions Q(0) = Q_o
- Carbon isotope reaches its half life in t = 5730 yrs
Find:
The expression for Q(t).
Solution:
- Assuming Q(t) satisfies:
Q' = -r*Q
- Separate variables:
dQ / Q = -r .dt
- Integrate both sides:
Ln(Q) = -r*t + C
- Make the relation for Q:
Q = C*e^(-r*t)
- Using initial conditions given:
Q(0) = Q_o
Q_o = C*e^(-r*0)
C = Q_o
- The relation is:
Q(t) = Q_o*e^(-r*t)
- We are also given that the half life of carbon is t = 5730 years:
Q_o / 2 = Q_o*e^(-5730*r)
-Ln(0.5) = 5730*r
r = -Ln(0.5)/5730
r = 0.000120968
- Hence, our expression for Q(t) would be:
Q(t) = Q_o*e^(-0.000120968*t)
