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Aneli [31]
3 years ago
7

The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 1 2 3 4 . A rot

ation reflection dilation translation must be used to make the two polygons coincide. A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a reflection across the x-axis and a 90° clockwise rotation about point D a translation 2 units down and a 90° counterclockwise rotation about point D a 90° counterclockwise rotation about point D and a reflection across the x-axis a reflection across the x-axis and a 90° counterclockwise rotation about point D

Mathematics
1 answer:
MaRussiya [10]3 years ago
8 0
I found the corresponding image. Pls. see attachment.

<span>The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).

A rotation translation must be used to make the two polygons coincide.

A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D </span>

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Step-by-step explanation:

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Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
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Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

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Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

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S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

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t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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Answer:

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