hi <3
439.612, 439.261, 439.216, 439.126
hope this helps :)
Answer:

Step-by-step explanation:
We will use the standard formula for half-life:

Where A₀ is the initial amount, a is the half-life, d is the time for one half-life, and t is the number of days.
So, we will substitute 1/2 for a and 24.3 for d. This yields:

We have a 3 gram sample. Therefore, our initial amount A₀ is 3:

And we want to find the remaining sample after 10 days. So, t=10. Substitute:

Use a calculator. So:

Therefore, for a 3 gram sample of Phosphorous-32 with a half-life of 24.3 days, after 10 days, there will be approximately 2.255 grams remaining.
Answer:
5! and 6p3
Step-by-step explanation:
5! = 5*4*3*2*1 = 120
6p3= n!/(n-r)!
where n = 6
and r = 3
6p3 = 6!/(6-3)!
= 6*5*4*3*2*1/3*2*1
= 120
Answer
P(milk)= 8/17
P(not milk)=9/17
Because 8+9=17
hope this was helpful!!!!