We know that
Half-life is modeled by the formula
An=A0*(0.5)<span>^[t/h)]
where
An----------> </span>is the amount remaining after a time t
A0----------> is the initial quantity
t------------> is the time
h------------> is the half-life of the decaying quantity
in this problem
h=1601 years
A0=50 g
An=?
t=100 years
An=A0*(0.5)^[t/h)]---------> An=50*(0.5)^[100/1601)]-----> 47.88 gr
the answer is 47.88 g
This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.
If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"
I show an example below of what I mean.
In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26
<h3>
Answer: 2.26</h3>
Answer:
a. 80 students
b. 92 students
Step-by-step explanation:
Represent arts students with A and Dance students with D.
So, we have,
n(A) = 35
n(D) = 57
Required
Determine n(A or D)
Solving (a):
Here, we have:
n(A and D) = 12
n(A or D) is calculated as thus:
n(A or D) = n(A) + n(D) - n(A and D)
n(A or D) = 35 + 57 - 12
n(A or D) = 80
b. From the given details
n(A and D) = 0 because both students are not mixed up as in (a) above
Using the same formula as (a).
n(A or D) = n(A) + n(D) - n(A and D)
n(A or D) = 35 + 57 - 0
n(A or D) = 92
