Answer:
1/2
Step-by-step explanation:
AC=18 and A'C'=9
AB(x)=A'B'
18(x)=9
x=9/18
x=1/2
Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
It would be B. because it's just 7 times 1.5
Answer:
frac{21x^6y^5}{14x^2y^9}
Factor the number =\frac{7\cdot \:3x^6y^5}{14x^2y^9}
Factor the number 14=7. 2 =\frac{7\cdot \:3x^6y^5}{7\cdot \:2x^2y^9}
Cancel\:the\:common\:factor:}\:7 =\frac{3x^6y^5}{2x^2y^9}
Step-by-step explanation: