Answer: Attached.
Step-by-step explanation:
Either solve the equations directly, or graph them and look for the point of intersection.
Answer:

Step-by-step explanation:
Slope-intercept form: y = mx + b
Slope formula: 
Given points: (3, -7), (7, 2)
(3, -7) = (x1, y1)
(7, 2) = (x2, y2)
To write the equation in slope-intercept form, we need to find the slope(m) and the y-intercept(b) of the equation.
First, let's find the slope. To do this, input the given points into the slope formula:

Simplify:
2 - (-7) = 2 + 7 = 9
7 - 3 = 4

The slope is
.
To find the y-intercept, input the slope and one of the given points(in this example I'll use point (7, 2)) into the equation and solve for b:



The y-intercept is
.
Now that we know the slope and the y-intercept, we can write the equation:

Answer:
where is the question so that I can solve it
Answer:
150 x - 80 y + 50 - 50 x - 25 y + 20
<em><u>Arranging </u></em><em><u>like </u></em><em><u>terms </u></em>
150 x - 50 x -25 y - 80y + 50 + 20
100x - 105 y + 70
<h3>5 ( 20x - 21y + 14) </h3>
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:

And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221