What is the value of y in the solution to the following system of equations?
1 answer:
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
Therefore,
Option D is the correct answer.
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Answer:
- x = -1
Explanation:
5x + 2 = 3x
5x - 3x = -2
2x = -2
x =
x = -1
Hope this helps! ☺
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:
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:
The test statistic value is, 68.184.
Decision:
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.