The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
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How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>
Suppose that we have:
- Sample size n > 30
- Sample mean =
- Sample standard deviation = s
- Population standard deviation =
- Level of significance =
Then the confidence interval is obtained as
- Case 1: Population standard deviation is known
- Case 2: Population standard deviation is unknown.
For this case, we're given that:
- Sample size n = 90 > 30
- Sample mean = = 138
- Sample standard deviation = s = 34
- Level of significance = = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).
At this level of significance, the critical value of Z is: 
= ±1.645
Thus, we get:
![CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]](https://tex.z-dn.net/?f=CI%20%3D%20%5Coverline%7Bx%7D%20%5Cpm%20Z_%7B%5Calpha%20%2F2%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5CCI%20%3D%20138%20%5Cpm%201.645%5Ctimes%20%5Cdfrac%7B34%7D%7B%5Csqrt%7B90%7D%7D%5C%5C%5C%5CCI%20%5Capprox%20138%20%5Cpm%205.896%5C%5CCI%20%5Capprox%20%5B138%20-%205.896%2C%20138%20%2B%205.896%5D%5C%5CCI%20%5Capprox%20%5B132.104%2C%20143.896%5D%20%5Capprox%20%5B130.10%2C%20143.90%5D)
Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
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originally. Then you reduces the side lengths a so it will be
Just note that 40% is the same as 2/5.<span />
Answer:
Step-by-step explanation:
Given that,
Initial speed of a car, u = 60 mph = 26.82 m/s
Final speed of a car, v = 10 mph = 4.47 m/s
Time, t = 5 s
We need to find the acceleration of the car. The rate of change of velocity is called the acceleration of an object. It can be given by :
So, the acceleration of the car is 
and it is deaccelerating.
Answer
when you place a source of signal at a focal point of an ellipse, the signal will be reflected to the other focal point position
When you place a source of rays as a signal at the focal point of a hyperbola, the ray will be reflected directly away from the other focal point
When you place a source of rays as a signal at one of the parabola focus, the ray will be reflected in lines that are parallel to the axis of the parabola.
L = 16.51
Get variable onto one side of equation and simplify:
(−7.4 )+ L = 9.11
L = 9.11 + 7.4
L = 16.51
