Ur output value is f(x).....so to do this problem u need to sub in 2/3 for f(x) and solve for the input value which is x
f(x) = -1/3x + 7
2/3 = -1/3x + 7 ...multiply everything by 3
2 = -x + 21
2 - 21 = - x
- 19 = -x
19 = x <== ur output
well, the sequence goes, 1100, to 1135, to 1170.... notice, is simply adding 35 to get the next term, so the common difference is 35, and the first term is 1100 of course.
![\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=1100\\ d=35\\ a_n=3725 \end{cases} \\\\\\ 3725=1100+(n-1)35\implies 3725=1100+35n-35 \\\\\\ 3725=1065+35n\implies 2660=35n\implies \cfrac{2660}{35}=n\implies 76=n](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%0A%5C%5C%5C%5C%0Aa_n%3Da_1%2B%28n-1%29d%5Cqquad%0A%5Cbegin%7Bcases%7D%0An%3Dn%5E%7Bth%7D%5C%20term%5C%5C%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ad%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa_1%3D1100%5C%5C%0Ad%3D35%5C%5C%0Aa_n%3D3725%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A3725%3D1100%2B%28n-1%2935%5Cimplies%203725%3D1100%2B35n-35%0A%5C%5C%5C%5C%5C%5C%0A3725%3D1065%2B35n%5Cimplies%202660%3D35n%5Cimplies%20%5Ccfrac%7B2660%7D%7B35%7D%3Dn%5Cimplies%2076%3Dn)
5000:100=50. 50*4=200 5000-200=4800 do the same 5times more, always taking the last number, so now start with 4800:100....
Answer:
$1.42 / day
Step-by-step explanation:
Divide 10 by 7. 10 for the amount of money being paid, 7 for the amount of days in a week.
Answer:
If a person selected at random from this group is a man, there is a 31.25% probability that the person favors raising the drinking age.
Step-by-step explanation:
We have these following informations:
There are 1000 people in total.
540 favored raising the age. Of them 390 were female, and 540-390 = 150 were male.
460 were opposed. Of them, 130 were female and 460-130 = 330 were males.
If a person selected at random from this group is a man, what is the probability that the person favors raising the drinking age?
There are 150+330 = 480 males. Of them, 150 favors raising the drinking age.
So

If a person selected at random from this group is a man, there is a 31.25% probability that the person favors raising the drinking age.