Use A = P(1+r/n)^(nt).
Here, A = $3000(1+0.07/12)^(12*6). Can you evaluate this?
Answer:
=18975
Step-by-step explanation:
This is an arithmetic sequence
an = a1 +d(n-1)
a1 =9
d = 12-9 = 3
an = 9 +3 (n-1)
The sum of an arithmetic sequence is given by
Sn = n/2 (a1+an)
So we need to determine what number in the sequence 336 is
an = 9 +3 (n-1)
336 = 9 +3(n-1)
Subtract 9 from each side
336-9 =9+-9 + 3(n-1)
327 = 3(n-1)
Divide each side by 3
327/3 =3/3(n-1)
109 = n-1
Add 1 to each side
109+1= n-1+1
110 = n
336 is the 110 term
The sum from 9 to 336
Sn = n/2 (a1+an)
S 110 = 110/2 (9+336)
= 55 (345)
=18975
I took the liberty of finding for the complete question.
And here I believe that the problem asks for the half life of Curium. Assuming
that the radioactive decay of Curium is of 1st order, therefore the
rate equation is in the form of:
A = Ao e^(-kt)
where,
A = amount after t years = 2755
Ao = initial amount = 3312
k = rate constant
t = number of years passed = 6
Therefore the rate constant is:
2755/3312 = e^(-6k)
-6k = ln (2755/3312)
k = 0.0307/yr
The half life, t’, can be calculated using the formula:
t’ = ln 2 / k
Substituting the value of k:
t’ = ln 2 / 0.0307
t’ = 22.586 years
or
t’ = 22.6 years
Answer:
It would be 16!
Step-by-step explanation:
Step 1 : 6 times 24 is 144
Step 2 : 144/9
Answer 16!
Answer:
c
Step-by-step explanation:
hope it helps
