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blagie [28]
3 years ago
6

In its 2016 Auto Reliability Survey, Consumer Reports asked subscribers to report their maintenance and repair costs. Most indiv

iduals 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 $170 and a sample of 24 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100.
Required:
State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.
Mathematics
1 answer:
lord [1]3 years ago
4 0

Answer:

See below

Step-by-step explanation:

Null Hypothesis: H_o:\sigma^2_1=\sigma^2_2, which means that the variance in annual repair costs is the same for both types of automobiles

Alternate Hypothesis: H_a:\sigma^2_1>\sigma^2_2, which means that the variance in annual repair costs is larger for older automobiles

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In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
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Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

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