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)
When we have a High Outlier in the data-set the line will extend from the Q3 to "Q3+1.5*IQR", which is considered the Maximum value, therefore the answer is True. The outlier will be a point outside this range.
A.4
hope this helped:)
plz mark brainliest
The answer is the square root of 207 or 20.2 ft(rounded to the nearest tenth)
Answer:
To find the inverse, you still interchange the variables
f - 1 (x) = 3 + x/2 - x is the Answer Hope this helps :)
Step-by-step explanation:
