What number is between 3/25 and √14? a) 1/14 b) 13/4 c) √20. d) 2 + π
2 answers:
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The solution of the linear equations will be ( -2, 4).
<h3>What is an equation?</h3>It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.
Given equations are:-
- 2x +3y = 8 and 3x+y= -2
Solving the equations by elimination method:-
2x +3y = 8
3x+y= -2
Multiply the second equation by 3 and subtract from the first equation.
2x +3y = 8
-9x -3y = 6
----------------
-7x = 14
x = -2
Out of the value of x in any one equation, we will get the value of y.
3x+y= -2
3 ( -2) + y = -2
-6 + y = -2
y = 4
The graph of the equations is also attached with the answer below.
Therefore the solution of the linear equations will be ( -2, 4).
The complete question is given below:-
Exploring Systems of Linear Equations 2x +3y =8 and 3x+y= -2. Find the value of x and y and draw a graph for the system of linear equations.
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Answer:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
(–2975, 175)
Do not reject H₀
Step-by-step explanation:
Hello!
The objective of this experiment is to test if the salaries of elementary school teachers are equal in two districts. You can test this trough the population means of the salaries of the teachers if they are either equal or different or directly test if the difference between the salaries of the two districts is cero or not, symbolically: μ₁ - μ₂ = 0
Remember, in the null hypothesis is usually stated the known information, is the "no change" premise and always carries the = symbol.
The hypothesis is:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
You have two normally distributed variables and you are studying the difference of the means. You can use a pooled Z to make the interval, the formula is:
X[bar]₁-X[bar]₂ ± *(√(σ₁²/n₁+σ₂²/n₂)
Since the test is two-tailed and at a signification level of 5% I've made the interval at the complementary confidence level of 95% so that I can use it to decide over the hypothesis.
X[bar]₁-X[bar]₂ ± *(√(σ₁²/n₁+σ₂²/n₂)
=
= 1,96
[28900-30300 ± 1.96*(√((2300)²/15+(2100)²/15)]
[-1400 ± 1576,15]
[-2976.15;176,15]
Since I've approximated to two decimal units in the intermediate calculations, the values differ slightly, but the interval is:
(–2975, 175)Now, since the 0 is contained in the Confidence interval, the decision is to not reject the null hypothesis. In other words, the difference in average salaries between the two districts is cero.
I hope it helps!