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Brrunno [24]
3 years ago
7

The F-ratio increases as _________. the variability between means increases relative to the variability within groups the variab

ility between means decreases relative to the variability within groups the total variability increases the total variability decreases
Mathematics
1 answer:
blondinia [14]3 years ago
6 0

Answer: variability between means increases relative to the variability within groups

Step-by-step explanation:

The formula to calculate F-ratio : variance between groups divided by variance within groups

Here, F-ratio \propto is proportional to variance between groups but inversely proportional to variance within groups .

So, F-ratio increases when variability between means increases relative to the variability within groups.

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What the answer to this equation X/8+4=5 ?
V125BC [204]

Answer:

The answer to this equation is 8

Step-by-step explanation:

\frac{x}{8} +4=5\\

<u>First: Subtract</u>

8-4=4

5 > 4 meaning that 5 is greater than 4.

You still need another number to make it equal than or greater than or equal to five.

5-4=1

Since 5 minus 4 does equal 1

You need somewhere to divide it from to get 1 so you can add up the 4 to it and get the answer to the equation.

<u>Lastly: Divide</u>

\frac{8}{8} =1

1+4=5\\5=5\\4\neq 5

Now, if you do 8/8 which equals 1, 1+4 = 5; Now everything is, <u>equaled</u> up!

<u>So in conclusion...</u>

\frac{x}{8} +4=5\\x=8\\\frac{8}{8}=1\\1+4=5\\5\neq5

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3 years ago
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Each letter of the alphabet is printed on an index card. What is the theoretical probability of randomly choosing any letter exc
kicyunya [14]
Probability=(number of specific outcomes)/(total number of possible outcomes)

P(!Z)=25/26  as a fraction exact

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The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon
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Answer and explanation:

Given : The position of an object moving along an x axis is given by x=3.24t-4.20t^2+1.07t^3 where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

x(t)=3.24t-4.20t^2+1.07t^3

x(1)=3.24(1)-4.20(1)^2+1.07(1)^3

x(1)=3.24-4.20+1.07

x(1)=0.11

b) At t= 2 s

x(t)=3.24t-4.20t^2+1.07t^3

x(2)=3.24(2)-4.20(2)^2+1.07(2)^3

x(2)=6.48-16.8+8.56

x(2)=-1.76

c) At t= 3 s

x(t)=3.24t-4.20t^2+1.07t^3

x(3)=3.24(3)-4.20(3)^2+1.07(3)^3

x(3)=9.72-37.8+28.89

x(3)=0.81

d) At t= 4 s

x(t)=3.24t-4.20t^2+1.07t^3

x(4)=3.24(4)-4.20(4)^2+1.07(4)^3

x(4)=12.96-67.2+68.48

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(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

\triangle x=x(4)-x(0)

\triangle x=14.24-0

\triangle x=14.24

(f) What is its average velocity from t = 2 s to t = 4 s?

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At t=4, x(4)=14.24

The average velocity  is given by,

\triangle x=x(4)-x(2)

\triangle x=14.24-(-1.76)

\triangle x=14.24+1.76

\triangle x=16

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