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2 years ago

PLEASE HELP!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

solong [7]2 years ago

6 0

Answer: True

In math terms, an "image" is the result after a transformation has been applied. One or more transformations can happen here. The original figure is called the "preimage" or "before image" so to speak.

So we go from "preimage" to "image"

preimage ---> image

A maps to A'

B maps to B'

C maps to C'

D maps to D'

zzz [600]2 years ago

3 0

Answer:

The answer is true

Step-by-step explanation:

I know because i had this problem.

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Solution :

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$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

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$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

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The grand mean or the overall mean is(GM) :

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$=\frac{51+51+...+49+49}{35}$

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The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

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The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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