Answer:
y = -1, x = -2
Step-by-step explanation:
First, multiply the top equation by 2 to make 3x become 6x:
6x - 8y = -4
Second, multiply the bottom equation by 3 to make -2x become -6x:
9y - 6x = 3
Add the two equations together, and you get:
y = -1
Plug in y as -1 into the first or second equation:
3x - 4(-1) = -2
OR
3(-1) - 2x = 1
x = -2
 
        
             
        
        
        
Answer:
The answer is explained below
Step-by-step explanation:
The question is not complete we need point P and point Q.
 let us assume P is at (3,1) and Q is at (-2,4)
To find the coordinate of the point that divides a line segment PQ with point P at  and point Q at
 and point Q at  in the proportion a:b, we use the formula:
 in the proportion a:b, we use the formula:

line segment PQ  is divided in the ratio 5:3 let us assume P is at (3,1) and Q is at (-2,4). Therefore:

 
        
             
        
        
        
Answer:
.14
Step-by-step explanation:
 
        
             
        
        
        
Answer:
4x + 5y = 20
Step-by-step explanation:
4x + 5y = 20
5y = -4x + 20      (Subtract 4x to isolate 5y. Put the 4x in front of the 20)
y =  + 4         (Divide all by 5 to isolate y)
 + 4         (Divide all by 5 to isolate y)
And then it's done and in slope-intercept form
 or
        or  is greater than -1 but less than 0
 is greater than -1 but less than 0
I hope this helped :)
 
        
             
        
        
        
Answer:
Step-by-step explanation:
From the given information:
The uniform distribution can be represented by:

The function of the insurance is:

Hence, the variance of the insurance can also be an account forum.
![Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2](https://tex.z-dn.net/?f=Var%20%5BI_%7B%28x%7D%29%20%3D%20E%20%5BI%5E2%28x%29%5D%20-%20%5BE%28I%28x%29%5D%5E2)
here;
![E[I(x)] = \int f_x(x) I (x) \ sx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cint%20f_x%28x%29%20I%20%28x%29%20%5C%20sx)
![E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cdfrac%7B1%7D%7B1500%7D%20%5Cint%20%5E%7B1500%7D_%7B250%7B%20%28x-%20250%29%20%5C%20dx)


Similarly;
![E[I^2(x)] = \int f_x(x) I^2 (x) \ sx](https://tex.z-dn.net/?f=E%5BI%5E2%28x%29%5D%20%3D%20%5Cint%20f_x%28x%29%20I%5E2%20%28x%29%20%5C%20sx)
![E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cdfrac%7B1%7D%7B1500%7D%20%5Cint%20%5E%7B1500%7D_%7B250%7B%20%28x-%20250%29%5E2%20%5C%20dx)


∴
![Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]](https://tex.z-dn.net/?f=Var%20%7BI%28x%29%7D%20%3D%201250%5E2%20%5CBig%20%5B%20%5Cdfrac%7B5%7D%7B18%7D%20-%20%5Cdfrac%7B25%7D%7B144%7D%5D)
Finally, the standard deviation  of the insurance payment is:


≅ 404