This is a combination problem.
Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.
The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.
This is the formula I used because the order is not important and repetition is not allowed.
Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.
numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960
Combination = 39,916,800 / 120,960 = 330
There are 330 ways that the instructor can choose 4 students for the first group
Given:
a varies jointly as b and c.
a=6 when b=2 and a=3.
To find:
The variation constant and the equation of variation.
Solution:
a varies jointly as b and c.

...(i)
Where, k is the constant of proportionality.
a=6 when b=2 and a=3.




The value of k is 1.
Putting k=1 in (i), we get


Therefore, the variation constant is 1 and the equation of variation is
.
The answer is b -3.6 hope this helps and god bless my friend
Answer:
-5/12
Step-by-step explanation:
Let's solve your equation step-by-step.
1
/3 = n + 3
/4
Step 1: Flip the equation.
n + 3
/4 = 1/
3
Step 2: Subtract 3/4 from both sides.
n+ 3
/4 − 3/4 = 1/
3 − 3
/4
n= −5
/12
The starting weight of the radioactive isotope = 96 grams
Weight after one hour is half of the starting weight. So the weight of the radioactive isotope after 1 hour = 48 grams
After 2 hours the weight is half as compared to the weight after previous hour. So weight after 2 hours = 24 grams.
This means, after every hour the weight is being halved. The half life of radioactive isotope is one hour.
Since after every hour, the weight is being halved, the weight of the isotope can be modeled by an exponential equation.
So,
Initial Weight = W₁ = 96
Change factor = 1/2 = 0.5
The general equation of the sequence will be:

Here t represents the number of hours. Using various values of t we can find the weight of the radioactive isotope at that time.
We can plot the sequence using the above equation. The graph is attached below.