An arithmetic sequence has a common difference.
143 - 130 = 13
156 - 143 = 13
169 - 156 = 13
The common difference is 13.
a1 = 130
a2 = 130 + 13
a3 = 130 + 2 * 13
a4 = 130 + 3 * 13
...
an = 130 + (n - 1) * 13
an = 130 + 13(n - 1)
an = 130 + 13n - 13
an = 117 + 13n
an = 13n + 117
Answer:
0.1505 = 15.05% probability that the hockey team wins 6 games in November
Step-by-step explanation:
For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use 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 that a certain hockey team will win any given game is 0.3723
So 
12 games in November
So 
What is the probability that the hockey team wins 6 games in November?
This is 


0.1505 = 15.05% probability that the hockey team wins 6 games in November
The verbal expression is "<em>Four enn-cubed plus six</em>."
Answer:
y-6
Step-by-step explanation: