Speeds
cutwrigth =35/84 =5/12 laps/min =0.417
evans =42/96.6=10/23 laps/min =0.435
loza =38/102.6 =10/27 laps/min =0.370
evans drove fastest
Answer: 
Step-by-step explanation:
Given
Length of the pipe 
Inside diameter of the pipe 
Outside diameter of the pipe 
Volume of the pipe
![\Rightarrow V=\dfrac{\pi }{4}[d_o^2-d_i^2]\\\\\text{Insert the values}\\\\\Rightarrow V=\dfrac{\pi}{4}[5^2-4^2]\times 10^{-4}\times 4\\\\\Rightarrow V=28.278\times 10^{-4}\ m^3\\\\\Rightarrow V=0.0028278\ m^3](https://tex.z-dn.net/?f=%5CRightarrow%20V%3D%5Cdfrac%7B%5Cpi%20%7D%7B4%7D%5Bd_o%5E2-d_i%5E2%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20V%3D%5Cdfrac%7B%5Cpi%7D%7B4%7D%5B5%5E2-4%5E2%5D%5Ctimes%2010%5E%7B-4%7D%5Ctimes%204%5C%5C%5C%5C%5CRightarrow%20V%3D28.278%5Ctimes%2010%5E%7B-4%7D%5C%20m%5E3%5C%5C%5C%5C%5CRightarrow%20V%3D0.0028278%5C%20m%5E3)
If all the dogs weighed the same, 44 would be the mean. The weight would be 44*5=220.
Hope this helped☺☺
12)
First of all, you need to know that profit is the difference between selling price and cost price. In this problem, profit is equal to the markup. Total profit is equal to the profit on each car multiplied by the number of cars sold.
The profit on each car is 2000 -400n, where n is defined in the problem statement as the number of markdowns (of $400).
The number of cars sold is 20 +6n. (Starting at 20, increasing by 6 for every markdown.)
The total monthly profit is the product of these binomials.
.. P(n) = (2000 -400n)*(20 +6n)
.. P(n) = -2400n^2 +4000n +40,000 . . . . . . selection C
13)
Matching functions to data points is often just a matter of trying them to see which fits. The first two function both match points (0, 2) and (1, 6). Functions C and D fail to match (1, 6).
Only the second function matches (2, 18).
The appropriate choice is
.. B. y = 2(3)^x