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zalisa [80]
3 years ago
8

An amusement park studied methods for decreasing the waiting time (minutes) for rides by loading and unloading riders more effic

iently. Two alternative loading/unloading methods have been proposed. To account for potential differences due to the type of ride and the possible interaction between the method of loading and unloading and the type of ride, a factorial experiment was designed. Use the following data to test for any significant effect due to the loading and unloading method, the type of ride, and interaction. Use α=α= .05.
Method 1 Method 2
Roller Coaster41434951
Type of RideScreaming Demon 52445046
Log Flume50464844
Method 1 Method 2
Roller Coaster
41
43
49
51
Type of Ride
Screaming Demon
52
44
50
46
​Log Flume
50
46
48
44

Mathematics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

hello your question has some missing part and the table as it relates better to the question

set up the ANOVA table ( to 1 decimal if necessary )

Step-by-step explanation:

attached below is the image of the Analysis of variance using Minitab software

Attached below as well is the Require Anova table

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Ray Of Light [21]

Answer:

Combine the terms with the same variable

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6 0
3 years ago
Let f be defined by the function f(x) = 1/(x^2+9)
riadik2000 [5.3K]

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

converges by comparison to the convergent <em>p</em>-series,

\displaystyle\sum_{n=3}^\infty\frac1{n^2}

(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

converges absolutely, since

\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0
3 years ago
If the balance of supplies at the start of the month was $900 and at the end of the month there was $450 on hand, the adjustment
Westkost [7]
A.450  If the balance of supplies at the start of the month was $900 then at the end of the month it was $450 so the answer would be 450.

Got this answer on weegy you should check it out sometimes 
4 0
4 years ago
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How are positive numbers written in Roman
rosijanka [135]

The system of Roman numerals is based on using special symbols to denote decimal digits I=1, X=10, C=100, M=1000, and their halves V=5, L=50, D=500. Positive integers are written by repeating these numerals.

 Brainliest?

5 0
3 years ago
-the equations x-y=-14 and -x-y=14 what is the number of solutions? A.) no solutions B.) infinitely many solutions C.) one solut
Yuki888 [10]

Answer:

The answer is C one solution

Step-by-step explanation:

x-y=-14

-y=-x-14  bring x to the other side

y=x+14   divide everything by -1

-x-y=14

-y=x+14  bring x to the other side

y=-x-14  divide everything by -1



6 0
3 years ago
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