6.59 written as a fraction would either be 659/100 or 6 59/100
3 < 5 if that is what you are contemplating
Answer:
B
Step-by-step explanation:
Stopping at a stoplight means that Chris's speed would be zero on the graph. The times that his speed is zero are times t = 0, t = 2 min, t = 11 min, and t = 21 min. Because the graph shows his speed from the time he left his house (at t = 0) to the time he arrived at the theater (at t = 21 min), the most likely times he stopped at stoplights are times t = 2 min and t = 11 min.
We are given with the set <span>K with members: 2, 4, 2, 0, 6, 0, 10, 8. The correct expression of set here is { 0 , 2 , 4 , 6, 8, 10 }. The numbers repeated are represented by one number in the set where the members are also arranged in order. </span>
Answer: 0.0250
Step-by-step explanation:
Given : A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 50.0 and 60.0 minutes.
We know that the uniform distribution is also known as rectangular distribution.
Density function for uniform distribution :

Now, the probability that a given class period runs between 51.25 and 51.5 minutes :-
![\int^{51.5}_{51.25}f(x)\ d(x)\\\\ \int^{51.5}_{51.25}\dfrac{1}{10}\ dx\\\\=\dfrac{1}{10} [x]^{51.5}_{51.25}\\\\=\dfrac{1}{10}(51.5-51.25)=0.0250](https://tex.z-dn.net/?f=%5Cint%5E%7B51.5%7D_%7B51.25%7Df%28x%29%5C%20d%28x%29%5C%5C%5C%5C%20%5Cint%5E%7B51.5%7D_%7B51.25%7D%5Cdfrac%7B1%7D%7B10%7D%5C%20dx%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B10%7D%20%5Bx%5D%5E%7B51.5%7D_%7B51.25%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B10%7D%2851.5-51.25%29%3D0.0250)
Hence, the probability that a given class period runs between 51.25 and 51.5 minutes = 0.0250