Answer:
(-3,-2)
Step-by-step explanation:
y = -x - 5
y = 2x + 4
plug in one of the y equations
(2x+4)= -x - 5
2x+4=-x-5
3x=-9
x= -3
plug in x to one of the y equations
y= -(-3) -5
y=3-5
y= -2
x= -3, y= -2
Answer: 
Step-by-step explanation:
First, we need to find the common denominator
The easiest way to do this is by multiplying the two given denominators, which are 2 and 5.
2 × 5 = 10
So, our common denominator is 10. Then, multiply the numerators of the two fractions by 2 and 5. Here's why:
=
We multiplied the top and bottom by 2 to make sure our new fraction stays equivalent to the original fraction.
Do the same thing for the other one:
= 
Finally, subtract the two fractions to find the difference between the two times:
= 
The reason we used the common denominator is because we can only add or subtract fractions if they have the same denominator.
I used a graphing calculator for this.
The solution my calculator got was:
(-2.625, 0.688) My calculator is very accurate, I don't think the test expects it to be like that.
Let's find the closest coordinates.
"B" is the closest.
I hope this helps!
~kaikers
Answer:
a) No
b) 42%
c) 8%
d) X 0 1 2
P(X) 42% 50% 8%
e) 0.62
Step-by-step explanation:
a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.
b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6
P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7
P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
c) P(win first game) = 0.4
P(win second game) = 0.2
P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%
d) X 0 1 2
P(X) 42% 50% 8%
P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%
e) The expected value 
f) Variance 
Standard deviation 