A radioactive substance is decaying exponentially. If there are initially 15 grams of the substance and after 3 hours there are 5 grams, what is the approximate half life of the substance, in hours? Round to two decimal places.

Ok, an exponential decay is written as:

P(t) = A*(1 - r)^t

Where A is the initial population, r is the rate of decay and t is the unit of time.

We know that the initial population is 15g

then:

P(t) = 15g*(1 - r)^t

And at t = 3hs, the populations is 5g

P(5h) = 5g = 15g*(1 - r)^5

5/15 = 1/3 = (1 - r)^5

(1/3)^(1/5) = (1 - r) = 0.8

Now, the half life time of the sustance is t = x, such that the population reduces to it's half:

P(x) = A/2 = 15g/2 = 7.5g

Then:

7.5g = 15g*0.8^x

7.5g/15g = 1/2 = 0.8^x

Now, remember that if we have:

a = b^x

then

x = ln(a)/ln(b)

ln(1/2)/ln(0.8) = x = 3,11 hours

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

read more on -

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)