Answer:
,, yes the confidence interval only contains positive values
Answer:you diddnt attach a graph but it woul have a stating pooint of positive 3 and a slove of up one over 1 so 1/1 +3
Step-by-step explanation:
This is a classic problem I think that you mean ![4^{n}+6n-1](https://tex.z-dn.net/?f=4%5E%7Bn%7D%2B6n-1)
Answer:
This problem clearly states to use proof by induction, follow the complete answer below.
Step-by-step explanation:
To prove something by induction we have to make a proposition first, in this case:
is divisible by 9
Where
is the order of the proposition.
First we have to prove the first few orders, let's just check P(1):
which is clearly divisible by 9
Now let's assume that P(n) is true that means, our original proposition is true. Now let's try to find out whether P(n+1) is true:
![4^{n+1}+6(n+1)-1=4\cdot 4^n+6n+5=4\cdot n+\underbrace{4\cdot 6n-18n}_{6n}-\underbrace{4\cdot 1+9}_{5}=4(4^n+6n-1)-18n+9=4(4^n+6n-1)-9(2n-1)](https://tex.z-dn.net/?f=4%5E%7Bn%2B1%7D%2B6%28n%2B1%29-1%3D4%5Ccdot%204%5En%2B6n%2B5%3D4%5Ccdot%20n%2B%5Cunderbrace%7B4%5Ccdot%206n-18n%7D_%7B6n%7D-%5Cunderbrace%7B4%5Ccdot%201%2B9%7D_%7B5%7D%3D4%284%5En%2B6n-1%29-18n%2B9%3D4%284%5En%2B6n-1%29-9%282n-1%29)
The
is divisible by 9 because we assumed that "
is divisible by 9" was a true proposition. On the other hand
has a factor of 9 regardless of n, thus is also divisible by 9 that means that P(n+1) is true and it follows that P(n) must also be true, thus proving our initial statement.
B is the answer to your question