If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
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A) x + y = 24
B) x^2 + y^2 = 306
A) x = 24 -y
Then substituting this into B)
(24 - y)^2 +y^2 = 306
576 -48y +y^2 + y^2 = 306
2 y^2 -48y + 270 = 0
x1 = 15
x2 = 9
Answer:

In which x is the number of which we want to find the probability.
Step-by-step explanation:
For each traffic fatality, there are only two possible outcomes. EIther it involved an intoxicated or alcohol-impaired driver or nonoccupant, or it didn't. Traffic fatalities are independent of other traffic fatalities, which means that the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
The probability is .40 that a traffic fatality involves an intoxication or alcohol-impaired driver or nonoccupant.
This means that 
Eight traffic fatalities
This means that 
Find the probability that the number which involve an intoxicated or alcohol-impaired driver or nonoccupant is
This is P(X = x), in which x is the number of which we want to find the probability. So

