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3 years ago

8

Side ST = 5, side US = 8, side VW = 5, and side XV = 8. What side corresponds to side TU and can be used to show that ?STU ? ?VW

X by SSS? (Enter your answer using letters only)

1 answer:

Debora [2.8K]3 years ago

5 0

If two sides of one triangle are congruent to two sides of another triangle then CPCTC makes it SSS

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Premises: All good students are good readers. Some math students are good students. Conclusion: Some math students are good read

just olya [345]

Answer:

Therefore, the conclusion is valid.

The required diagram is shown below:

Step-by-step explanation:

Consider the provided statement.

Premises: All good students are good readers. Some math students are good students.

Conclusion: Some math students are good readers.

It is given that All good students are good readers, that means all good students are the subset of good readers.

Now, it is given that some math students are good students, that means there exist some math student who are good students as well as good reader.

Therefore, the conclusion is valid.

The required diagram is shown below:

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A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari

guajiro [1.7K]

Answer:

a

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

b

$F_{critical} = 1.8608$

c

$F = 2.9085$

d

The decision rule is

Reject the null hypothesis

e

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 30$

The sample variance for elementary school is $s^2_1 = 8324$

The second sample size is $n_2 = 30$

The sample variance for the secondary school is $s^2_2 = 2862$

The significance level is $\alpha = 0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

Generally from the F statistics table the critical value of $\alpha = 0.05$ at first and second degree of freedom $df_1 = n_1 - 1 = 30 - 1 = 29$ and $df_2 = n_2 - 1 = 30 - 1 = 29$ is

$F_{critical} = 1.8608$

Generally the test statistics is mathematically represented as

$F = \frac{s_1^2 }{s_2^2}$

=> $F = \frac{8324 }{2862}$

=> $F = 2.9085$

Generally from the value obtained we see that $F > F_{critical }$ Hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to support the researchers claim

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Answer:

Step-by-step explanation:

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