To simplify the expression:
-(6 x^3 - 2 x + 3) - 3 x^3 + 5 x^2 + 4 x - 7
Start with - (6 x^3 - 2 x + 3) = -6 x^3 + 2 x - 3:
-6 x^3 + 2 x - 3 - 3 x^3 + 5 x^2 + 4 x - 7
Next, combine similar terms: -3 x^3 - 6 x^3 + 5 x^2 + 4 x + 2 x - 7 - 3 = (-3 x^3 - 6 x^3) + 5 x^2 + (4 x + 2 x) + (-7 - 3):
(-3 x^3 - 6 x^3) + 5 x^2 + (4 x + 2 x) + (-7 - 3)
-3 x^3 - 6 x^3 results in -9 x^3:
-9 x^3 + 5 x^2 + (4 x + 2 x) + (-7 - 3)
Combine 4 x and 2 x to get 6 x:
-9 x^3 + 5 x^2 + 6 x + (-7 - 3)
The operation -7 - 3 yields -10:
-9 x^3 + 5 x^2 + 6 x - 10
Factoring out -1 from -9 x^3 + 5 x^2 + 6 x - 10 leads to:
Final Answer: - (9 x^3 - 5 x^2 - 6 x + 10)
1
2
3
Step-by-step explanation: Generally, during the roll of two fair 6-sided dice, the doubles result in (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Therefore, the total for doubles is N = 6. The outcome of rolling two fair 6-sided dice yields n = 36. Thus, the probability of rolling doubles (matching numbers on both dice) is calculated mathematically. When rolling two fair dice, outcomes that sum to 4 or less are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Observing this, we see two doubles present. Consequently, the conditional probability of rolling doubles is represented mathematically. Lastly, when rolling the two fair dice, outcomes that show different numbers result in L = 30, while outcomes where at least one die shows a 1 give W = 10. Hence, the conditional probability of having at least one die show a 1 is presented mathematically.
Answer:
Step-by-step explanation:
Given that lines l and m are parallel and there is a transversal cutting through these lines.
5). (9x + 2)° = 119° [by alternate interior angles]
9x = 117 ⇒ x = 13
6). (12x - 8)° + 104° = 180°
12x = 180 - 96
x = 
⇒ x = 7
7). (5x + 7) = (8x - 71) [by alternate exterior angles]
8x - 5x = 71 + 7
3x = 78
x = 26
8). (4x - 7) = (7x - 61) [by corresponding angles]
7x - 4x = -7 + 61
3x = 54
x = 18
9). (9x + 25) = (13x - 19) [by corresponding angles]
13x - 9x = 25 + 19
4x = 44
x = 11
(13x - 19)° + (17y + 5)° = 180° [linear pairs of angles sum to supplementary]
(13×11) - 19 + 17y + 5 = 180
129 + 17y = 180
17y = 180 - 129
y = 3
10). (3x - 29) + (8y + 17) = 180 [linear pairs of angles sum to supplementary]
3x + 8y = 180 + 12
3x + 8y = 192 -----(1)
(8y + 17) = (6x - 7) [by alternate exterior angles]
6x - 8y = 24
3x - 4y = 12 -----(2)
From equation (1) subtract equation (2)
(3x + 8y) - (3x - 4y) = 192 - 12
12y = 180
y = 15
Plugging into equation (1),
3x + 8(15) = 192
3x + 120 = 192
x = 24
11). (3x + 49)° = (7x - 23)° [by corresponding angles]
7x - 3x = 49 + 23
4x = 72 ⇒ x = 18
(11y - 1)° = (3x)° [by corresponding angles]
11y = 3×18 + 1
11y = 55 ⇒ y = 5
12). (5x - 38)° = (3x - 4)° [by corresponding angles]
5x - 3x = 38 - 4
2x = 34
x = 17
(7y - 20)° + (5x - 38)° + 90° = 180°
[The angles within a triangle sum to 180°]
7y + 5x - 58 = 90
5x + 7y = 148
5×17 + 7y = 148
85 + 7y = 148
7y = 148 - 85
y = 
Answer:
E. P(W | H)
Step-by-step explanation:
Meaning of each probability:
P(H): Likelihood of the game occurring at home
P(W): Likelihood of the game resulting in a win.
P(H and W): Likelihood that the game is both at home and a win.
P(H|W): Likelihood of a win occurring when at home.
P(W|H): Likelihood of winning at home games.
Which probabilities are necessary to determine the fraction of home games that were victories?
This is the probability of winning a home game. Hence, the answer is:
E. P(W | H)
