There would be a a row of seven and would be 9 in each row
Answer:
28%
Step-by-step explanation:
So I am assuming the numbers are the number of students.
So 64 + 28+ 52 +56 = 200 students
Assuming the 56 outside the diagram are NOT attending either college.
so 56/200 =28%
Answer:
a) 2.9%
b) Option B is correct.
The prisoners must be independent with regard to recidivism.
Step-by-step explanation:
Probability that one prisoner goes back to prison = 17% = 0.17
a) The probability that two prisoners released both go back to prison = 0.17 × 0.17 = 0.0289 = 2.89% = 2.9% to 1 d.p
b) The only assumption taken during the calculation is that probability of one of the prisoners going back to prison has no effect whatsoever in the probability that another prisoner goes back to prison. That is the probability that theses two events occur are totally independent of each other.
If they weren't, we wouldn't be able to use 0.17 as the probability that the other prisoner goes back to prison too.
Answer is ''there are infinitely many intersection points'' .
Let equation of the line 1 be

As given they are the same line so Russ should interpret that

So, they are coincident and its a consistent-dependent system .Both of them will have same values of y for same value of x .
Therefore they will have infinitely many solutions and hence intersections.
For example Equation of line 1 be 2x +3y = 5 and equation of line 2 be 4x+6y=10 then
therefore, these lines are coincident and have infinitely many solutions .
There is only 20 different ways you can do for 1 topping and 1 cheese for this question. Because every topping(10) on every cheese(2) gives you 20 different choices you can have a pizza.