The 99% Confidence interval for the mean sale time for all homes in the neighborhood is:87.857, 112.143.
<h3>Confidence interval</h3>
a. The assumption is: Assume that the population has a normal distribution.
The CI is exact for the normal populations and for small samples the z-interval method should be used in a situation where the variable is normally distributed.
b. Confidence interval:
CI=Sample mean±z-score×Standard deviation/√Size of the sample
CI=100-2.576×20/√18, 100+2.576×20/√18
CI=100-12.143, 100+12.143
CI=87.857, 112.143
Therefore the 99% Confidence interval is 87.857, 112.143.
Learn more about Confidence interval here:brainly.com/question/15712887
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The answer is B 1 =52 2=66
Answer:
5 hours
Step-by-step explanation:
We know that Joey spent a total of $325. He first paid the fee of $150. Let's subtract $150 from $325.
$325 - $150 = $175
Now, we know that it costs $35 per hour to rent the car. So, we divide $175 by $35 to find out how many hours Joey rented the car for.
$175 ÷ $35 = 5 hours
Check the picture below.
let's check where the quadratic and x = 6 meet, bear in mind we're integrating by "y", so
![\bf -------------------------------\\\\ \displaystyle \int\limits_{0}^{\frac{\pi }{2}}\ [6-6sin(y)]dy\implies \left. \cfrac{}{} 3y^2+6cos(y) \right]_{0}^{\frac{\pi }{2}} \\\\\\ \left[ \cfrac{3\pi 2}{4}+0 \right]-[0+6]\implies \cfrac{3\pi^2}{4}-6\approx 1.4022033](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5C%20%5B6-6sin%28y%29%5Ddy%5Cimplies%20%5Cleft.%20%5Ccfrac%7B%7D%7B%7D%203y%5E2%2B6cos%28y%29%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft%5B%20%5Ccfrac%7B3%5Cpi%202%7D%7B4%7D%2B0%20%5Cright%5D-%5B0%2B6%5D%5Cimplies%20%5Ccfrac%7B3%5Cpi%5E2%7D%7B4%7D-6%5Capprox%201.4022033)
