A scientist left a 32-gram sample of a radioactive substance out and it was exposed to the air. After 30 minutes, only 26 grams remained.
How much was left when the janitor found it 8 hours after the scientist left?
What is the half-life of this substance?
Hey, can anyone help me solve this. I'm not sure i'm answering it correctly. Thank you.

There were 1.155 g of the substance eight hours later and it's halflife time is 1.66943 h.

Step-by-step explanation:

Since radiactive decay is expressed by a exponential function of the following kind:

mass(t) = mass(0)*e^(-a*t)

Where mass(t) is the mass of the substance at a given time, mass(0) is the inital mass substance, a is a factor that defines how fast the substance will decay and t is the elapsed time. The problem provided us with all the needed information to create the expression, with the exception of the "a". So we can apply it to the known point in order to determine "a". We have:

mass(0.5) = 32*e^(-a*0.5)

26 = 32*e^(-a*0.5)

e^(-0.5*a) = 26 / 32

e^(-0.5*a) = 0.8125

ln[e^(-0.5*a)] = ln(0.8125)

-0.5*a*ln(e) = ln(0.8125)

-0.5*a = ln(0.8125)

a = -ln(0.8125)/0.5 = 0.4152

So the expression that represents the mass of the substance over time is given by:

mass(t) = 32*e^(-0.4152*t)

After 8 hours we have:

mass(8) = 32*e^(-0.4152*8) = 32*e^(-3.3216) = 1.155 g

The half life is the time it'll take to reach half it's original mass, therefore:

16 = 32*e^(-0.4152*t)

32*e^(-0.4152*t) = 16

e^(-0.4152*t) = 16 / 32

e^(-0.4152*t) = 0.5

ln[e^(-0.4152*t)] = ln(0.5)

-0.4152*t = ln(0.5)

t = -ln(0.5)/0.4152 = 1.66943 h

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: