2X^2+4X-8-X^2+6X-4
X^2+4X-8+6X-4
X^2+10X-12
when you add two negatives its just like regular number, -3+-5=-8 same as 3+5=8
Answer:
-$13.5
Step-by-step explanation:
Let x be a random variable of a count of player gain.
- We are told that if the die shows 3, the player wins $45.
- there is a charge of $9 to play the game
If he wins, he gains; 45 - 9 = $36
If he looses, he has a net gain which is a loss = -$9
Thus, the x-values are; (36, -9)
Probability of getting a 3 which is a win is P(X) = 1/6 since there are 6 numbers on the dice and probability of getting any other number is P(X) = 5/6
Thus;
E(X) = Σ(x•P(X)) = (1/6)(36) + (5/6)(-9)
E(X) = (1/6)(36 - (5 × 9))
E(X) = (1/6)(36 - 45)
E(X) = -9/6 = -3/2
E(X) = -3/2
This represents -3/2 of $9 = -(3/2) × 9 = - 27/2 = -$13.5
the question does not present the options, but this does not interfere with the resolution
we know that
if a and b are parallel lines
so
1) m∠2=58°------> by corresponding angles
2) m∠1=4x-10------> by alternate exterior angles
3) [m∠2+m∠(3x-1)]+m ∠1=180°------> by supplementary angles
58+(3x-1)+4x-10=180
7x=180-47
7x=133
x=19°
4) angle (4x-10)=-4*19-10-------> 66°
5) angle 3x-1=3*19-1-------> 56°
How many models does the following set have? 5,5,5,7,8,12,12,12,150,150,150
Strike441 [17]
<h3>
Answer: 3 modes</h3>
The three modes are 5, 12, and 150 since they occur the most times and they tie one another in being the most frequent (each occurring 3 times).
Only the 7 and 8 occur once each, and aren't modes. Everything else is a mode. It's possible to have more than one mode and often we represent this as a set. So we'd say the mode is {5, 12, 150} where the order doesn't matter.
