This is a probability problem with two dependent events and conditional probability. Note that after the first donut is chosen, it is not replaced into the data set, so only 23 donuts remain. If we set A=selection of a lemon-filled, and B=selection of a custard-filled, then P(A and B) = P(A)*P(B|A), where P(B|A) means the probability of B happening given that A has already occurred.P(A) = 8/24 = 1/3 = 0.333333P(B|A) = 12/23 = 0.521739P(A and B) = 1/3(12/23) = 12/69 = 0.1739130435 or 17.4%
https://www.wyzant.com/resources/answers/296921/find_the_probability_of_selecting_a_a_lemon_filled_d...
Answer:
It's y=3x
Step-by-step explanation:
The y-intercept is 0, it's direct variation. The slope(rise/run) is 3.
If the numbering of the angles is similar to this:
1 | 2
-----
3 | 4
5 | 6
----
7 | 8
Then none of the the choices are necessarily true.
The equations would have been:
1 = 4 = 5 = 8
2 = 3 = 6 = 7
3 + 5 = 4 + 6 = 180
1. x= 8
2. x=2
3. x=-9
4. x=4
5. x= -5
6. x=16
7 x=0
8: x=-3
Hope this helps!
