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Stolb23 [73]
3 years ago
9

A researcher reports an F-ratio with df = 1, 24 for an independent-measures experiment. If all the treatments had the same numbe

r of participants, then how many individuals were in each treatment?​
Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

13 participants.

Step-by-step explanation:

Given that a researcher reports an F-ratio with df = 1, 24 for an independent-measures experiment.

i.e. degrees of treatment between groups =1

and within groups = 24

This implies that there are two groups.

For two groups to have same number of participants, since we get total of degrees of freedom =24+1 =25

Hence total items selected = 26

i.e. each group would have 13 entries

So each treatment had 13 participants.

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Step-by-step explanation:

y is the amount of money

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Miranda has earned scores of 20, 22, 18, and 24 on her last four math quizzes. What score will she need to earn on the fifth qui
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16

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4 0
3 years ago
40% of x is 35. What is the equation to find x
Dima020 [189]

Answer:

.40 x = 35  equation

x = 87.5

Step-by-step explanation:

Of means multiply and is means equals

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4 0
3 years ago
Read 2 more answers
Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2
LUCKY_DIMON [66]
Make a change of coordinates:

u(x,y)=xy
v(x,y)=\dfrac xy

The Jacobian for this transformation is

\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}

and has a determinant of

\det\mathbf J=-\dfrac{2x}y

Note that we need to use the Jacobian in the other direction; that is, we've computed

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}

but we need the Jacobian determinant for the reverse transformation (from (x,y) to (u,v). To do this, notice that

\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to

\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv
=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2
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3 years ago
A sequence of transformations is described below.
swat32

Answer:

I don't understand this

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