Answer:
hope it helps mark as brainliest
For the first problem, the answer is D, because every year, the graph goes down by about $4,500.
For problem two,
a. It's located in quadrant one because x and y are both positive (I've attached a graph with labeled quadrants for reference)
I'm unsure about b and c but I hope I helped with the others!
Answer:
P = 0.05
Step-by-step explanation:
12 months * 30 days each = 360 days
From 306 days, we have to select 8 days = 360C8 ways(Total ways)
We want each days from different month. First, we have to select 8 month from 12 month = 12C8 ways
---By selecting 8 month, we will select a days from each month. That can be done in = 30C1 * 30C1 * .................30C1 (8 ways) [From a month with 30 days, we can select a day in 30C1 ways = 30 ways]
Therefore P = Number of ways of selecting each days from different month / Total number of ways
P = 12C8 * 30^8 / 360C8
P = 495 * 656100000000 / 6469697679132645
P = 0.0501985588982791
P = 0.05
Hence the probability that each day is from a different month is 0.05
First, we subtract 12x from both sides. We then get -4y=-12x+48. We then divide this by -4 on both sides to get the y by itself. We then get that the equation is y=3x-12.
Answer:
16x^2 + 9x^2 + 9x + 13.
Step-by-step explanation:
6x^3 + 8x^2 – 2x + 4 +10x^3 + x^2 + 11x + 9
Bringing like terms together:
= 6x^3 + 10x^3 + 8x^2 + x^2 - 2x + 11x + 4 + 9
= 16x^2 + 9x^2 + 9x + 13. (answer).
