What are the solutions to the following system of equations?

It is given that trapezoid EFGH is an isosceles trapezoid. We know that FE ≅ GH by the definition of

jeyben [28]

Answer:

1. isosceles trapezoid

2. FEH

3. GHE

4. Reflexive

5. SAS

Step-by-step explanation:

Edg2020!!

3 0

3 years ago

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobil

o-na [289]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.

State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.

At a 0.01 level of significance, what is your conclusion? What is the p-value?

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.44

p-value = 0.1954

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

Step-by-step explanation:

Let σ₁² denotes the variance of 4 years old automobiles

Let σ₂² denotes the variance of 2 years old automobiles

State the null and alternative hypotheses:

The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic is given by

Test statistic = σ₁²/σ₂²

Test statistic = 120²/100²

Test statistic = 1.44

p-value:

The degree of freedom corresponding to 4 years old automobiles is given by

df₁ = n - 1

df₁ = 26 - 1

df₁ = 25

The degree of freedom corresponding to 2 years old automobiles is given by

df₂ = n - 1

df₂ = 23 - 1

df₂ = 22

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.44, 25, 22)

p-value = 0.1954

Conclusion:

When the p-value is less than the significance level then we reject the Null hypotheses

p-value < α (reject H₀)

But for the given case,

p-value > α

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

8 0

3 years ago

Which is a rational number?<br> 3.<br> 4.<br> 5/5<br> 6.12

k0ka [10]

Answer:

i pretty sure one of em is 6.12

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%282%20%5Cdiv%202%20%5Csqrt%7B2%29%20%5E%7B2%7D%20%7D%20" id="TexFormula1" title="(2 \div 2 \s

Genrish500 [490]

Hopes this helps:

Answer: 2

6 0

3 years ago

A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than

vovikov84 [41]

Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

$H_0: p \leq 0.05$

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

$H_a: p > 0.05$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that $\mu = 0.05, \sigma = \sqrt{0.05*0.95}$

A random sample of 300 items taken from the production line contained 27 defective items.

This means that $n = 300, X = \frac{27}{300} = 0.09$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}$

$z = 3.18$

Pvalue of the test:

Testing if the mean is greater than a value, which means that the pvalue of the test is 1 subtracted by the pvalue of Z = 3.18, which is the probability of a finding a sample proportion of 0.09 or higher.

Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

5 0

3 years ago