We can use point slope form to solve for this.
y - 1 = -2(x - 2)
Simplify.
y - 1 = -2x + 4
Add 1 to both sides.
y = -2x + 5
Now, we can input -3 for x and solve for y, or r in this case.
y = -2(-3) + 5
y = 6 + 5
y = 11
r = 11
<span>35 - 10 ➗ 5 + [(5 + 3) • 4]
Do PEMDAS
1.52</span>
I’m pretty sure but not so sure I think it is 180
Answer: C) y ≥ 3x - 2; 
<u>Step-by-step explanation:</u>
Blue line:
y-intercept (b) = -2
slope (m) is 3 up, 1 right = 3
shading is above
⇒ y ≥ 3x - 2
Yellow line:
y-intercept (b) = 3
slope (m) is 1 up, 2 right = 
shading is below

Answer:
Does this sample provide convincing evidence that the machine is working properly?
Yes.
Step-by-step explanation:
<u>Normal distribution test:</u>

Where,







Once the significance level was not given, It is usually taken an assumption of a 5% significance level.
Taking the significance level of 5%, which means a confidence level of 95%, we have a z-value of 
Therefore, we <u>fail to reject the null</u>. It means that the hypothesis test is not statistically significant: the average length is not different from 1.5!