Sample space for:
First shooting a basketball. Two options: X or O
Then rolling a die. Six options: 1, 2, 3, 4, 5, or 6
Sample space:
X1
X2
X3
X4
X5
X6
O1
O2
O3
O4
O5
O6
Answer: Options:
A. X1
C. O6
F. X6
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
alright whats your question ?
9514 1404 393
Answer:
32
Step-by-step explanation:
The first term is 5 and the common difference is 9-5=4, so the n-th term is ...
an = a1 +d(n -1)
an = 5 +4(n -1) = 4n +1
Then the term number for the term 129 is ...
129 = 4n +1 . . . . . put the given value in the formula
128 = 4n . . . . . . . subtract 1
32 = n . . . . . . . . . .divide by 4
The 32nd term is 129.
Step-by-step explanation:
given :
2x - 3y = 11
-6x + 8y = 34
find : the solutions of the system by using Cramers Rule.
solutions:
in the matrix 2x2 form =>
[ 2 -3] [x] [11]
=
[-6 8] [ y] [34]
D =
| 2 -3 |
|-6 8 |
= 8×2 - (-3) (-6)
= 16-18 = -2
Dx = | 11 -3 |
| 34 8 |
= 11×8 - (-3) (34)
= 88 + 102
= 190
Dy = | 2 11 |
|-6 34 |
= 2×34 - (-6) (11)
= 68 + 66
= 134
x = Dx/D = 190/-2 = -95
y = Dy/D = 134/-2 = -67
the solutions = {-95, -67}