First you have to figure out how much a video game costs and how much a used one costs
Then you plug in the costs of the video games into Janets equation 120=3x+y
(x= video games and y=used video games)
Then you subtract the cost of video games from 120 and then divide that answer by the cost of used video games and that should give you how many used video games she can get

3 0

3 years ago

Helppppppppppppppppppppppppppppppppppp

ELEN [110]

Answer:

Step-by-step explanation:

18/6=3

8 0

3 years ago

Every day, a comes to the bus stop at the same time, and boards the first bus that arrives. the arrival of the first bus is an e

mylen [45]

So the waiting time for a bus has density f(t)=λe−λtf(t)=λe−λt, where λλ is the rate. To understand the rate, you know that f(t)dtf(t)dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.

So define X=min(B1,B2)X=min(B1,B2), where B1B1 is exponential with rate 33 and B2B2 has rate 44. It's easy to show the minimum of two independent exponentials is another exponential with rate λ1+λ2λ1+λ2. So you want:

P(X>20 minutes)=P(X>1/3)=1−F(1/3),P(X>20 minutes)=P(X>1/3)=1−F(1/3),

where F(t)=1−e−t(λ1+λ2)F(t)=1−e−t(λ1+λ2).

8 0

3 years ago