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ozzi
3 years ago
8

What is the value of y in the solution to the following system of equations?

Mathematics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

Option D is the correct answer.

Step-by-step explanation:

The equations are:

5x - 3y = -11     Eqn 1

2x - 6y = -14     Eqn 2

We have to find the value of y so we will eliminate x.

Multiplying Eqn 1 by 2

2(5x-3y=-11)

10x-6y=-22     Eqn 3

Multiplying Eqn 2 by 5

5(2x-6y=-14)

10x-30y=-70     Eqn 4

Subtracting Eqn 4 from Eqn 3

(10x-6y)-(10x-30y)= -22-(-70)

10x-6y-10x+30y = -22 + 70

24y = 48

\frac{24y}{24}=\frac{48}{24}\\y=2

Therefore,

Option D is the correct answer.

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Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

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The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

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Stels [109]

Answer:

30

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(-) (-) = +

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