Help me please!!!!!!!
1 answer:
You might be interested in
Answer:
The equation in point-slope form is:
Step-by-step explanation:
We need to find equation of line in point slope form that passes through points (5,0) and (-6, -1).
The general equation of point slope form is:
where m is the slope of line.
We need to know Slope m
Finding Slope m
Finding Slope m using formula:
We have
Putting values and finding slope
So , slope m is:
Using point (5,0) and slope m =1/22 the equation of point-slope form is:
So, the equation in point-slope form is:
Answer:
Confidence interval: (1760,1956)
Step-by-step explanation:
We are given the following information in the question:
Sample size, n = 81
Sample mean =
Population standard deviation =
Confidence Level = 95%
Significance level = 5% = 0.05
Confidence interval:
Putting the values, we get,
Step-by-step explanation:
Q3
(a) If 0 < |x − 1| < δ, then |f(x) − 2| < 2
What this means is, how far can x stray from the x=1 line such that f(x) stays within 2 units of the y=2 line (0 < f(x) < 4).
If we move 2 units left of x=1, we get f(x) = 4.
If we move about 3.5 units right of x=1, we get f(x) = 4.
Therefore, δ can't be more than 2.
(b) If 0 < |x| < δ, then |f(x) − 3| < 1
What this means is, how far can x stray from the x=0 line such that f(x) stays within 1 unit of the y=3 line (2 < f(x) < 4).
If we move 1 unit left of x=0, we get f(x) = 4.
If we move 1 unts right of x=0, we get f(x) = 2.
Therefore, δ can't be more than 1.
Since f(x) isn't continuous within this domain, we can't conclude that the limit exists.
Q4
(a) Yes. If δ = 0.25, then 0.75 < x < 1.25, and f(x) > 200.
(b) No. f(1) = 300, so even if δ = 0, f(x) will be less than 400.
(c) Yes. If δ ≈ 0.1, then 0.9 < x < 1, and f(x) > 450.
