For this case we have that by definition, the equation of a line in the slope-intercept form is given by:

Where:
m: Is the slope
b: Is the cut-off point with the y axis
We have the following equation:

We manipulate algebraically:
We subtract 10 from both sides of the equation:

We subtract 3x from both sides of the equation:

We multiply by -1 on both sides of the equation:

We divide between 5 on both sides of the equation:

Thus, the equation in the slope-intercept form is 
Answer:

Answer:
3 cups of water
Step-by-step explanation:
Answer:
A. 2 ( 10 x − 5 )
D. 10 ( 2 x − 1 )
Step-by-step explanation:
Given expression:
20x − 10
Find the equivalent expression
Check all that applies
A. 2 ( 10 x − 5 )
= 20x - 10
B. 2 ( 2 x − 1 )
= 4x - 2
C. 10 ( 20 x − 10 )
= 200x - 100
D. 10 ( 2 x − 1 )
20x - 10
The correct options are
A. 2 ( 10 x − 5 )
D. 10 ( 2 x − 1 )
Answer:

Step-by-step explanation:
For a uniform distribution we have to:
for x∈ (a, b)
for 
for 
In this case we have to:
a = 25 min and b = 105 min
We want to find:

Then


In this case:

Therefore:


