This one bc there’s one point vertically and there’s no more than one point when you draw a vertical line down
Answer:
Incomplete question, but you can use the formulas given to solve it.
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
![P(X < x) = \frac{x - a}{b - a}](https://tex.z-dn.net/?f=P%28X%20%3C%20x%29%20%3D%20%5Cfrac%7Bx%20-%20a%7D%7Bb%20-%20a%7D)
The probability of finding a value between c and d is:
![P(c \leq X \leq d) = \frac{d - c}{b - a}](https://tex.z-dn.net/?f=P%28c%20%5Cleq%20X%20%5Cleq%20d%29%20%3D%20%5Cfrac%7Bd%20-%20c%7D%7Bb%20-%20a%7D)
The probability of finding a value above x is:
![P(X > x) = \frac{b - x}{b - a}](https://tex.z-dn.net/?f=P%28X%20%3E%20x%29%20%3D%20%5Cfrac%7Bb%20-%20x%7D%7Bb%20-%20a%7D)
Uniform distribution over an interval from 0 to 0.5 milliseconds
This means that ![a = 0, b = 0.5](https://tex.z-dn.net/?f=a%20%3D%200%2C%20b%20%3D%200.5)
Determine the probability that the interarrival time between two particles will be:
Considering
, and the question asked, you choose one of the three formulas above.
Answer:
![y = 4 \left(x - \dfrac{9}{8} \right)^2 - \dfrac{65}{16}](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28x%20-%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20-%20%5Cdfrac%7B65%7D%7B16%7D%20)
Step-by-step explanation:
![y = 4x^2 - 9x + 1](https://tex.z-dn.net/?f=%20y%20%3D%204x%5E2%20-%209x%20%2B%201%20)
You need to complete the square on the right side.
![y = (4x^2 - 9x) + 1](https://tex.z-dn.net/?f=%20y%20%3D%20%284x%5E2%20-%209x%29%20%2B%201%20)
![y = 4(x^2 - \dfrac{9}{4}x) + 1](https://tex.z-dn.net/?f=%20y%20%3D%204%28x%5E2%20-%20%5Cdfrac%7B9%7D%7B4%7Dx%29%20%2B%201%20)
![y = 4 \left( x^2 - \dfrac{9}{4}x + \left( \dfrac{9}{8} \right)^2 \right) + 1 - 4\left( \dfrac{9}{8} \right)^2](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28%20x%5E2%20-%20%5Cdfrac%7B9%7D%7B4%7Dx%20%2B%20%5Cleft%28%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20%5Cright%29%20%2B%201%20-%204%5Cleft%28%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20)
![y = 4 \left(x - \dfrac{9}{8} \right)^2 + 1 - 4 \left( \dfrac{81}{64} \right)](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28x%20-%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20%2B%201%20-%204%20%5Cleft%28%20%5Cdfrac%7B81%7D%7B64%7D%20%5Cright%29%20)
![y = 4 \left(x - \dfrac{9}{8} \right)^2 + 1 - \dfrac{81}{16}](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28x%20-%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20%2B%201%20-%20%5Cdfrac%7B81%7D%7B16%7D%20)
![y = 4 \left(x - \dfrac{9}{8} \right)^2 + \dfrac{16}{16} - \dfrac{81}{16}](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28x%20-%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20%2B%20%5Cdfrac%7B16%7D%7B16%7D%20-%20%5Cdfrac%7B81%7D%7B16%7D%20)
![y = 4 \left(x - \dfrac{9}{8} \right)^2 - \dfrac{65}{16}](https://tex.z-dn.net/?f=%20y%20%3D%204%20%5Cleft%28x%20-%20%5Cdfrac%7B9%7D%7B8%7D%20%5Cright%29%5E2%20-%20%5Cdfrac%7B65%7D%7B16%7D%20)
Answer:
see explanation
Step-by-step explanation:
Under a reflection in the y- axis
a point (x, y ) → (- x, y ) , then
J (1, 4 ) → J' (- 1, 4 )
K (1, 0 ) → K' (- 1, 0 )
L (4, 3 ) → L' (- 4, 3 )
Plot the points J', K', L' and connect them in order to obtain image