
we know all it's doing is adding 6 over again to each term to get the next one, so then

now for the explicit one
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D7%5C%5C%20d%3D6%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D7%2B%28n-1%296%5Cimplies%20a_n%3D7%2B6n-6%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BExplicit%20Formula%7D%7D%7B%5Cstackrel%7Bf%28n%29%7D%7Ba_n%7D%3D6n%2B1%7D%20%5C%5C%5C%5C%5C%5C%20therefore%5Cqquad%20%5Cqquad%20f%2810%29%3D6%2810%29%2B1%5Cimplies%20f%2810%29%3D61)
Answer:
(-4,6)
Step-by-step explanation:
y=6
5x+5(6)=10
5x+30=10
5x=-20
x=-4
(x,y)
(-4,6)
It is A so you have it right
Answer:
11
Step-by-step explanation:
(Note that the diagram is not to scale)
You know BC is 6
You know DC is 5
You know CE is 12
Using K's position, it shows that AC is one off from EC
AC=11
Answer:
0.0082 = 0.82% probability that he will pass
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. 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.
In this problem we have that:
.
If the student makes knowledgeable guesses, what is the probability that he will pass?
He needs to guess at least 9 answers correctly. So









0.0082 = 0.82% probability that he will pass