Let

be a probability that i-th light fails, where i=1,2,3. Then,

is the probability that i-th light remain bright.
<span>If the lights fail independently of each other, the probability that a string of lights will remain bright for two years is:
</span>

.
-4(2.5y-3.25) - 15.25 = -10y + box
-4(-0.75y) - 15.25 = -10y + box
3y-15.25 = -10y + box
-7y-15.25 = box
box and/or y = -22.25
i'm not 100 percent sure of my answer, but I'm pretty confident. If i am wrong, then all you need to do is distribute in the beginning.
Answer:
*a general admission ticket is $160
*a grandstand ticket is $390
Explanation: make 2 equations from the provided information
**g is a grandstand ticket and a is a general admission ticket.
4g+2a=1880, 4g+4a=2200
Steps: 4g+2a=1880 changes to 4g=1880-2a because you move the 2a in order to solve for g.
You can substitute the 4g into 4g+4a=2200 to get 1880-2a+4a=2200 and simplify to get 1880+2a=2200. subtract 1880 from both sides and you get 2a=320. divide both sides by 2 to get a=160.
After that you can substitute the value for a into 4g+4a=2200 to get 4g+4(160)=2200. simplify to get 4g+640=2200. subtract 640 from both sides to get 4g=1,560. divide both sides by 4 to get g=390.
Hope this helped! :)
Answer:
(4,-2)
Step-by-step explanation:
If we are reflecting a point across the y-axis, that point must be on the same horizontal line that the pre-mage is was, this means the y coordinate must remain the same. Now we want the points on both sides of the y-axis to have the same distance. x and -x have the same distance from 0. This means you will need to take the opposite of x.
So a point (x,y) being reflected over the y-axis will give (-x,y).
( x , y)->(-x,y) means we have:
(-4,-2 )->(4,-2)
You can graph this to confirm it looks like a reflection through the y-axis.