25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Answer:
791.68 cm/s
Step-by-step explanation:
The volume flow rate can be interpreted as the integral of fluid velocity over area
![\dot{V} = \int\limits^6_0 {v(r) 2\pi r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {(25-r^2)r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {25r-r^3} \, dr\\\\\dot{V} = 2\pi[12.5r^2 - r^4/4]_0^6\\\dot{V} = 2\pi(12.5*6^2 - 6^4/4 - 12.5*0 - 0)\\\dot{V} = 2\pi*126 = 791.68 cm/s](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%20%3D%20%5Cint%5Climits%5E6_0%20%7Bv%28r%29%202%5Cpi%20r%7D%20%5C%2C%20dr%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5Cint%5Climits%5E6_0%20%7B%2825-r%5E2%29r%7D%20%5C%2C%20dr%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5Cint%5Climits%5E6_0%20%7B25r-r%5E3%7D%20%5C%2C%20dr%5C%5C%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5B12.5r%5E2%20-%20r%5E4%2F4%5D_0%5E6%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%2812.5%2A6%5E2%20-%206%5E4%2F4%20-%2012.5%2A0%20-%200%29%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%2A126%20%3D%20791.68%20cm%2Fs)
Y=(3x+4)(-2x-3)=-6x^2-4x-9x-12=-6x^2-13x-12.
It is a quadratic function because it has x in power 2 (x^2)
NO it is not possible when dividing an irregular shape to polygons to overlap. Every polygon would be broken up in piece.
Answer:
a question was smarter
Step-by-step explanation:
but we cant help u with the scatter plot use desomos
