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VladimirAG [237]
3 years ago
13

Anyone have any idea what it could be?

Mathematics
1 answer:
Ronch [10]3 years ago
4 0
The answer is B because 90 to 30 is 3 over 2
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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
16. Interest earned:
photoshop1234 [79]

\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$1250\\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ t=years\dotfill &4 \end{cases} \\\\\\ I=(1250)(0.03)(4)\implies I=150

3 0
3 years ago
Suppose that the useful life of a car battery, measured in months, decays with parameter 0.03 . We are interested in the life of
scZoUnD [109]

Answer:

a) X ~ e(0.03)

b) μ = 100/3

c) \sigma = 100/3

d) A battery is expected to last 100/3 months (33 months and 10 days approximately).

e) For seven batteries, i would expect them to last 700/3 months (approximately 19 years, 5 months and 10 days).

Step-by-step explanation:

a) The life of a battery is usually modeled with an exponential distribution X ~ e(0.03)

b) The mean of X is μ = 1/0.03 = 100/3

c) The standard deviation is \sigma = 1/0.03 = 100/3

d) The expected value of the bateery life is equal to its mean, hence it is 100/3 months.

e) The expected value of 7 (independent) batteries is the sum of the expected values of each one, hence it is 7*100/3 = 700/3 months.

4 0
3 years ago
Evaluate the expresssion |–12|–|8|.
ruslelena [56]
This essentially means 12-8
|| means absolute value which is the distance from 0 the number is
Basically it’s the number but positive

So the answer is 4
5 0
3 years ago
Read 2 more answers
A business owner has just paid $6000 for a computer. It depreciates at a rate of 22% per year. How much will it be worth in 5 ye
katovenus [111]

Answer:

It will be worth $1732.30.

Step-by-step explanation:

After 1 year : 6000 x .78 = 4680

2 yr - 4680 x .78 = 3650.4

3 yr - 3650.4 x .78 = 2847.312

4 yr - 2847.312 x .78 = 2220.90336

5 yr - 2220.90336 x .78 = 1732.304621

1732.30

I think this is what it was asking

7 0
3 years ago
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