Answer:
First number is -2.5
Second number 5
Step-by-step explanation:
Four times the first number minus three times the second number is 2.
To represent this statement, 4x -3y =2
Two times the first number increased by three times the second is 10.
To represent thsi statement, 2x+ 3y = 10
4x - 3y = 2
2x + 3y =10
4x - 2x = 2x
2x = -5
x= -2.5
The first number is -2.5
By using Substitution
2x + 3y = 10
2(-2.5) + 3y = 10
-5 + 3y = 10
3y = 10 +5
3y =15
y = 5
The second number is 5
....To check my answer
-5 + 15 = 10
Idk tbh wat is a great game for me so I can play with the kids and play it for a while and then play me and play me when I get home I wanna Play is time
simplifying
we get 
Step-by-step explanation:
We need to simplify: 
Solving:
Applying the rule:
log a + log b = log(ab)

So, simplifying
we get 
Keywords: Simplifying Logarithms
Learn more about Simplifying Logarithms at:
#learnwithBrainly
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