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Vika [28.1K]
4 years ago
13

Tony believes that the more time he spends at the record store, the more money he tends to spend. To see if he is correct, he ke

pt a record of his recent purchases each time he visited the record store and displayed them in the scatter plot below. He also determined the line of best fit.
The equation of the line is y=2.4 x - 1.8
record store visit
Based on the graph shown, which of the following statements are true?
Select THREE that apply.

A
The line of best fit will have a positive correlation coefficient.

B
Each additional minute Tony spends in the record store is associated with an additional cost of $2.40.

C
The correlation coefficient for the line of best fit is 2.4.

D
If Tony stays for 30 minutes in the record store, it is likely he will spend $70.

Mathematics
1 answer:
mixer [17]4 years ago
4 0

Answer:

A, B, D

Step-by-step explanation:

It's not C because the correlation coeffecient connot be greater than 1 or less than -1.

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Find the mass of the solid paraboloid Dequals=​{(r,thetaθ​,z): 0less than or equals≤zless than or equals≤8181minus−r2​, 0less th
Lubov Fominskaja [6]

Answer:

M = 5742π  

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

                             ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

                             0 ≤ z ≤ 81 - r^2

                             0 ≤ r ≤ 9

Find:-

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Solution:-

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                           M = \int \int \int {p ( r ,theta, z)} \, dV\\\\

- We can write the above expression in cylindrical coordinates:

                           M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\

- Perform integration:

                           M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\

                           M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] }  \, dtheta\\\\M = 5742\pi  kg              

- The mass evaluated is M = 5742π                      

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