Answer:
c would be the answer 31/ 3
Math pretty much has its own language. In an expression, u have to " break it apart " to make each term
one more then the product of a number and 7 decreased by 5...
" product " means multiply.....so the product of a number and 7 would be 7n.
one more then the product....(7n + 1)....." decreased" means subtract....
so ur expression is : (7n + 1) - 5
(7n + 1) - 5....when n = 6
(7(6) + 1) - 5 =
43 - 5 =
38 <==
Answer:
3
Step-by-step explanation:
i think
Check the picture below, so the circle looks more or less like so, with a radius of 9.
![\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=9 \end{cases}\implies C=2\pi (9)\implies C\approx 57](https://tex.z-dn.net/?f=%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D9%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%20%289%29%5Cimplies%20C%5Capprox%2057)
The expected value of the first game is -$0.50 and of the second game is -$0.52.
There are 10³ possible numbers for the lottery, and only 1 of them will match in the correct order; this gives a probability of 1/1000. To find the expected value, we multiply this by the winnings (499 after the $1 cost); we also multiply the probability of losing (999/1000) by the amount lost (-1):
1/1000(499)+999/1000(-1)
499/1000 - 999/1000 = -500/1000 = -0.50
For the second game, since the number is "boxed", there are 3! ways to get the correct digits; this gives a probability of 6/1000. Multiply this by the winnings, 79 (after the $1 cost); multiply the probability of losing (994/1000) by the loss (-1):
6/1000(79) + 994/1000(-1) = 474/1000 - 994/1000 = -520/1000= -0.52