Levi purchased a total of 50 ride and game tickets and the amusement park and spent $29. if ride tickets cost 75 cents each and
game tickets cost 50 cents each, how many ride tickets did he purchase?
2 answers:
Hello there!
Your question states: Levi purchased a total of 50 ride and game tickets and the amusement park and spent $29. if ride tickets cost 75 cents each and game tickets cost 50 cents each, how many ride tickets did he purchase?
x = rides
y = games
<span>x + y= 50
<span>0.75x + 0.50y = 29
To solve this system of equations, we can graph (attached below).
As we can see, the answer is (16, 34)
This means that he purchased 16 ride tickets and 34 game tickets.
Hope I helped!
let me know if you need anything else!
~ Zoe</span></span>
Your question states: Levi purchased a total of 50 ride and game tickets and the amusement park and spent $29. if ride tickets cost 75 cents each and game tickets cost 50 cents each, how many ride tickets did he purchase?
x = rides
y = games
<span>x + y= 50
<span>0.75x + 0.50y = 29
To solve this system of equations, we can graph (attached below).
As we can see, the answer is (16, 34)
This means that he purchased 16 ride tickets and 34 game tickets.
Hope I helped!
let me know if you need anything else!
~ Zoe</span></span>
You might be interested in
Answer: 0.5898
Step-by-step explanation:
Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .
We assume that,
The probability that .J. Redick makes any given free throw =0.901 (1)
Free throws are independent.
So it is a binomial distribution .
Using binomial probability formula, the probability of getting success in x trials :
, where n= total trials
p= probability of getting in each trial.
Let x be binomial variable that represents the number of a=makes.
n= 14
p= 0.901 (from (1))
The probability that he makes at least 13 of them will be :-
∴ The required probability = 0.5898