Answer:
<em>Two possible answers below</em>
Step-by-step explanation:
<u>Probability and Sets</u>
We are given two sets: Students that play basketball and students that play baseball.
It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.
This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

P = 0.66
Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:
We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.
Thus, there 19-2=17 students who play only one of the sports. The probability is:

P = 0.59
4y ≥ 3x + 2
Plugging in the values in the options, then the required values are when x = 6 and y = 5, then
4(5) ≥ 3(6) + 2
20 ≥ 18 + 2
20 ≥ 20
The best way to solve this is by giving a visual idea of the glass ornaments. so We will have letter G represent Green, O for Orange, B for Brown, P for Purple.
Sample 1:G,O,B,P
Sample 2:G,P,O,B
Sample 3:P,G,O,B
Sample 4:O,G,P,B
Sample 5:G,O,P,B
This is what you can do this should be the right amount of ways to arrange them.
Answer:
A)
x= the number of ride tickets
y= the total cost of admission plus how many ride tickets a person purchases
B)
y= 1.25x + 9.5
C)
It is a$1.25 per ticket for the rides at the fair, so it would be 1.25 multiplied by the amount of tickets that are purchased (x). Spencer bought 17 tickets, so 17x1.25= 21.25 and it says that he spent a total of $30.75 at the fair, so 30.75-21.25=9.5, so that means the cost of admission is $9.50.
Step-by-step explanation:
I hope this helps!
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