Answer:
19.3 years
Step-by-step explanation:
Given that the initial mass of a sample of Element X is 100 grams,
The formula is given as:
N(t) = No × (1/2) ^t/t½
Element X is a radioactive isotope such that every 30 years, its mass decreases by half.
N(t) = Mass after time (t)
No = Initial mass = 100 grams
t½ = Half life = 30 grams
N(t) = 100 × (1/2) ^t/30
How long would it be until the mass of the sample reached 64 grams, to the nearest tenth of a year?
This means we are to find the time
N(t) = 100 × (1/2) ^t/30
N(t) = 64 grams
64 = 100(1/2)^t/30
Divide both sides by 100
64/100 = 100(1/2)^t/30/100
0.64 = (1/2)^t/30
Take the Log of both sides
log 0.64 = log (1/2)^t/30
log 0.64 = t/30(1/2)
t = 19.315685693242 years
Approximately = 19.3 years
In short and without much fuss
let's say Anne puts "x" amount in the account at 1.2% rate annually, that means after 1 year, she will have "x" + 1.2% of "x", or 1.012x to be exact.
the 1.2% rate, kicks in as the period of a year is met.
now, what if Anne puts it in the monthly compounded type? that means, the compounding period is a month, so after 1 month, she has 1.2% extra, or 1.012x, and after 2 months, she will have 1.2% extra of 1.012x, or 1.012144x, and after 3 months, she will have 1.2% extra of 1.0121x, or 1.012145728x and so on.
anyhow, the shorter the compounding period, the more the 1.2% kicks in, the more accumulation in the account.
This is a combination since the order doesn't matter
C(15, 6) = (15 * 14 * 13 * 12 * 11 * 10)/(6 * 5 * 4 * 3 * 2 * 1)
C(15, 6) = 3603600/720
C(15, 6) = 5005 different combinations of votes
