This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to cone back to the skipped part. The half-life of cesium-137 is 30 years. Suppose we have a 30 mg sample. Exercise
(a) Find the mass that remains after t years. Step 1 Let y(t) be the mass (in mg) remaining after t years. Then we know the following Stop 2 Since the half-life is 30 years, then y(30) - More Information m Your answer cannot be understood or graded Submit Skie (you cannot come back) Exercise
(b) How much of the sample remains after 20 years? Step 1 After 20 years we have the following (20) 30 mg (Round your answer to two decimal places.)

(a)y(t)=30exp(-0.0231t)

(b)y(20)=18.9mg

Step-by-step explanation:

At a particular time t, the mass of a radioactive substance like Cesium-137 is governed by the equation:

N=N₀e⁻ᵏᵗ where k=ln 2/half life

(a)Mass that remains after t years

Half Life= 30 years

k= ln2/30=0.0231

Initial Mass, N₀=30mg

Therefore the mass N that remains at time t

N=N₀e⁻ᵏᵗ

N=30exp(-0.0231t)

y(t)=30exp(-0.0231t)

(b)We want to determine how much of the sample remains after 20 years.

At t=20 years

y(t)=30exp(-0.0231t)

y(20)=30exp(-0.0231X20)

=30 X 0.63

y(20)=18.9mg

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,