Answer: ![f(x)=1\ \text{where}\ 12\leq x\leq 13](https://tex.z-dn.net/?f=f%28x%29%3D1%5C%20%5Ctext%7Bwhere%7D%5C%2012%5Cleq%20x%5Cleq%2013)
Step-by-step explanation:
The probability distribution function for uniform distribution in interval [a,b] is given by :_
![f(x)=\dfrac{1}{b-a}\ \text{where}\ a\leq x\leq b](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B1%7D%7Bb-a%7D%5C%20%5Ctext%7Bwhere%7D%5C%20a%5Cleq%20x%5Cleq%20b)
Given : Mr. Warren Buffet and Mr. Zhao Danyang agree to meet at a specified place between 12 pm and 1 pm.
Since 1 pm comes after 12 pm, and in 24 hours system we call it as 13 pm.
If each person arrives between 12 pm and 1 pm at random with uniform probability, then the uniform distribution function for this will be :_
![f(x)=\dfrac{1}{13-12}=1\ \text{where}\ 1\leq x\leq 12](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B1%7D%7B13-12%7D%3D1%5C%20%5Ctext%7Bwhere%7D%5C%201%5Cleq%20x%5Cleq%2012)
Hence. the distribution function for the length of the time that the first to arrive has to wait for the other will be :-
![f(x)=1\ \text{where}\ 12\leq x\leq 13](https://tex.z-dn.net/?f=f%28x%29%3D1%5C%20%5Ctext%7Bwhere%7D%5C%2012%5Cleq%20x%5Cleq%2013)
(2/3) of the box is done in (1/2) minute.
The 1 whole fraction of the box will be done in (1/2) ÷ (2/3) = 1/2 × 3/2 = 3/4 minutes
1 box is packed in (3/4) minutes.
Number of boxes per minute = (1 box) / (3/4) minute = 4/3 boxes per minute.
Answer:
49$ on Friday night
105$ on Saturday night
77$ Sunday Afternoon
$231 during the weekend
Step-by-step explanation:
7$ perdog just mean multiplication so
7*7=49
7*15=105
7*11=77
Now for the whole weekend we have to add so 49+105+77= 231$ in total
2×10^(-7)=0.0000002m
You have 6 zeros in front of 2.