Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays to 356g over 7 days.

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

The half life of the sample of phosphorus-32 is

Step-by-step explanation:

The equation used to calculate the half life of the sample is given as:

where,

m(t) =  amount of sample after time 't' = 356 g

= initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

Hence, the half life of the sample of phosphorus-32 is

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

your answer is -9

step-by-step explanation: