Answer:
The 95% confidence interval for the true population mean of pollutant discharge is (26.7 tons, 33.7 tons).
Step-by-step explanation:
Let <em>X</em> represent the daily level of pollutant discharge in the manufacturing district in Pittsburgh.
The monitoring equipment reveals that for the 28 days of the February month
Mean daily discharge per factory, () = 30.2 tons
Standard deviation daily discharge per factory, (<em>s</em>) = 9.1 tons
Compute the critical value of <em>t</em> for 95% level of confidence and (n - 1 =) 27 degrees of freedom as follows:
*Use a t-table.
Compute the 95% confidence interval for the true population mean of pollutant discharge as follows:
Thus, the 95% confidence interval for the true population mean of pollutant discharge is (26.7 tons, 33.7 tons).
Answer:
Step-by-step explanation:
Area of a Segment in Radians A = (½) × r2 (θ – Sin θ)
Area of a Segment in Degrees A = (½) × r 2 × [(π/180) θ – sin θ]
Answer:
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Answer:
ok
Step-by-step explanation:
ok
1 yard^2 = 0.8361m^2.?
<span>The solution to the problem is as follows:
So he spent (.625 x 400) = $250.00 per m^2. </span>
He paid out (500 x 250) = $125,000 on the deal.
He received (500 / 0.8361) = 598.014592 yard^2 of fabric.
That would have cost (598.014592 x 120) = $71,761.75 locally.
He lost (125,000 - 71,761.75) = $53,238.25 on the deal.
<span>You can rework with the incorrect conversion factor.
I hope my answer has come to your help. God bless and have a nice day ahead!
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