Answer:
12.1%
Step-by-step explanation:
Given that:
Mean (μ) = 20.2 grams and standard deviation (σ) = 0.18 grams.
The z score is a score used to determine the number of standard deviations by which the raw score is above or below the mean. A positive z score means that the raw score is above the mean and a negative z score means that the raw score is below the mean. It is given by:
a) For x < 19.99 g:
From the normal distribution table, P(x < 19.99) = P(z < -1.17) = 0.1210 = 12.1%
The probability that a randomly chosen mouse has a mass of less than 19.99 grams is 12.1%
You set (x+15) equal to 40 and solve for x.
x+15=40
-15 -15
x=25
Hope this helps!
Answer:
E
Step-by-step explanation:
Sjsjdjddjdjdjjddjdjdjdjdjdndnsnsnsbs answer is E
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)
