Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
We are given the <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:
- Sample mean of
![\overline{x} = 12.17](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%3D%2012.17)
- Population standard deviation of
![\sigma = 6.37](https://tex.z-dn.net/?f=%5Csigma%20%3D%206.37)
- Sample size of
.
The margin of error is:
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which z is the critical value.
We have to find the critical value, which is z with a p-value of
, in which
is the confidence level.
In this problem,
, thus, z with a p-value of
, which means that it is z = 1.96.
Then:
![M = 1.96\frac{6.37}{\sqrt{50}} = 1.77](https://tex.z-dn.net/?f=M%20%3D%201.96%5Cfrac%7B6.37%7D%7B%5Csqrt%7B50%7D%7D%20%3D%201.77)
The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:
![\overline{x} - M = 12.17 - 1.77 = 10.4](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20-%20M%20%3D%2012.17%20-%201.77%20%3D%2010.4)
![\overline{x} + M = 12.17 + 1.77 = 13.94](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%2B%20M%20%3D%2012.17%20%2B%201.77%20%3D%2013.94)
The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
A similar problem is given at brainly.com/question/22596713
the formula is y=mx+b
m is the slope
b is the y-intercept form
so the -1 is the y-intercept form. so you need to start at that point on the y axis. then the slope which is -1/2 comes into play. in my school we use rise/run. so from the point you marked at -1 you will go down 1 and to the right 2 points then where you stop is your next point (mark it) keep doing that. also go up the other way from the -1 point. go up 1 and to the left 2. then connect all of your points with a line. I hope this helped.
please chose this answer as brainliest
Answer:
53 lies between 7.2² and 7.3²
Step-by-step explanation:
Estimating a root to the nearest tenth can be done a number of ways. The method shown here is to identify the tenths whose squares bracket the value of interest.
You have answered the questions of parts 1 to 3.
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<h3>4.</h3>
You are given that ...
7.2² = 51.84
7.3² = 53.29
This means 53 lies between 7.2² and 7.3², so √53 lies between 7.2 and 7.3.
53 is closer to 7.3², so √53 will be closer to 7.3 than to 7.2.
7.3 is a good estimate of √53 to the tenths place.
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<em>Additional comment</em>
For an integer n that is the sum of a perfect square (s²) and a remainder (r), the square root is between ...
s +r/(2s+1) < √n < s +r/(2s)
For n = 53 = 7² +4, this means ...
7 +4/15 < √53 < 7 +4/14
7.267 < √53 < 7.286
Either way, √53 ≈ 7.3.
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The root is actually equal to the continued fraction ...
![\sqrt{n}=\sqrt{s^2+r}=s+\cfrac{r}{2s+\cfrac{r}{2s+\dots}}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%3D%5Csqrt%7Bs%5E2%2Br%7D%3Ds%2B%5Ccfrac%7Br%7D%7B2s%2B%5Ccfrac%7Br%7D%7B2s%2B%5Cdots%7D%7D)
X = 4 and Y = 9. I did a complicated way of doing it so if u copy it it would look weird