Answer:choices?
Step-by-step explanation:
I hope this helps you
n!/(n-3)!=3.n!/(n-2)!
3. (n-3)!=(n-2)(n-3)!
3=n-2
n=5
Each portion will weigh 0.079lbs.
5/6-1/5=?
.833-.2=.633
.633÷8=.079lbs.
Answer:
that is simplified as much as possible
Answer:
0.81 = 81% probability that a randomly selected student is taking a math class or an English class.
0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class
Step-by-step explanation:
We solve this question working with the probabilities as Venn sets.
I am going to say that:
Event A: Taking a math class.
Event B: Taking an English class.
77% of students are taking a math class
This means that 
74% of student are taking an English class
This means that 
70% of students are taking both
This means that 
Find the probability that a randomly selected student is taking a math class or an English class.
This is
, which is given by:

So

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.
Find the probability that a randomly selected student is taking neither a math class nor an English class.
This is

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class