Functions cannot have the same X value (the first number), but they can have the same Y value (the second number).
<span>A. {(1,2),(2,3),(3,4),(2,1),(1,0)}
B. {(2,−8),(6,4),(−3,9),(2,0),(−5,3)}
C. {(1,−3),(1,−1),(1,1),(1,3),(1,5)}
D. {(−2,5),(7,5),(−4,0),(3,1),(0,−6)}
Choice A. has two repeating X values [(1,2) and (1,0), (2,3) and (2,1)]
Choice B. has one repeating X value [(2, -8) and (2,0)]
Choice C. all has a repeating X value (1)
Choice D doesn't have any repeating X values.
In short, your answer would be choice D [</span><span>{(−2,5),(7,5),(−4,0),(3,1),(0,−6)}] because it does not have any repeating X values.</span>
Answer:
2/7
Step-by-step explanation:
There are 22 (12 + 10) total students in the class. That means that the chance of the first student picked being a girl is 12/22.
Now, we must calculate the chance of the next student to be picked <em>also </em>being a girl - however, there is a trap here! Remember that since a girl has been picked, the total student pool has decreased to 21 and and the total number of girls has decreased to 11. This means the new chance of girl being picked is 11/21.
To find the probability of both these events happening in conjunction, these fractions must be multiplied. 12/22 * 11/21 = 132/462, which simplifies to 2/7.
If there are 3 people in the family and each ate 2 slices. 6 slices were eaten in total.
To find the ammount of pizza eaten, divide the total slices eaten (6) over the original ammount of slices (8)
Eaten ammount = 6/8 = 3/4
Answer:
1 - 1/10
Step-by-step explanation:
As strange as it may seem, the easiest way to answer this question is to calculate the probability of selecting an elephant first and then to subtract this probability from 1:
Adding up the Numbers of animals, we get 12+4+3+11, or 30. The probability of selecting an elephant is thus 3/30, or 1/10.
Thus, the probability of NOT selecting an elephant is 1 - 1/10 (Answer A).
