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Triss [41]
2 years ago
5

HELP ME PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS URGENTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Mathematics
1 answer:
bazaltina [42]2 years ago
5 0

9514 1404 393

Answer:

  90 m²

Step-by-step explanation:

The given linear dimensions have the ratio ...

  red/blue = (6 m)/(4 m) = 3/2

The ratio of areas is the square of the ratio of the linear dimensions:

  (red area)/(blue area) = (3/2)²

  S = (blue area)(9/4) = (40 m²)(9/4)

  S = 90 m²

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Answer:a lighter, my girl, and a gun

Step-by-step explanation:

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1/6 times what is 4<br> please answer asap
Scorpion4ik [409]
Answer is 24, becuase if you just do the opposite, 4 divided by 1/6 you get 24
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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
Please answer!
Xelga [282]

Amswer

it is b Step-by-step explanation:

8 0
2 years ago
(55 - 52 ) + (+3 + 6)<br> Order of operations
True [87]

Answer: 12

Step-by-step explanation:

(55-52) + ( 3 +6)

3 + 9

12

6 0
3 years ago
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