Q(a+y)=67y+93 Solve for y
2 answers:
<h2>Answer:</h2>
The value of y is:
We are given a expression in terms of three unknowns q,a and y.
Now, we are asked to solve this expression for y.
i.e. we are asked to represent y in terms of other two unknowns.
We will first use the distributive property under multiplication in the left hand side of the equation.
i.e.
Now, we combine the terms of y at one side of the equation and the other terms to the other side.
i.e.
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Answer:
Part A
b. 14.6 ± 7.38
Part B
b. 3.43
Part C
a. P-value < 0.01
Part D
b. There is sufficient evidence to reject the null hypothesis
Step-by-step explanation:
Part A
The given data are;
The number of seedlings in the field = 20
The number of seedlings selected to receive herbicide A = 10
The number of seedlings selected to receive herbicide B = 10
The height in centimeters of seedlings treated with Herbicide A, = 94.5 cm
The standard deviation, s₁ = 10 cm
The height in centimeters of seedlings treated with Herbicide B, = 109.1 cm
The standard deviation, s₂ = 9 cm
The 90% confidence interval for μ₂ - μ₁, is given as follows;
The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;
The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18
α = 100% - 90% = 10%
∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05
= 1.734
C.I. ≈ 14.6 ± 7.37714603353
The 90% C.I. ≈ 14.6 ± 7.38
b. 14.6 ± 7.38
Part B
With the hypotheses are given as follows;
H₀; μ₂ - μ₁ = 0
Hₐ; μ₂ - μ₁ ≠ 0
The two sample t-statistic is given as follows;
The two sample t-statistic ≈ 3.43
b. 3.43
Part C
From the t-table, the p-value, we have, the p-value < 0.01
a. P-value < 0.01
Part D
Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis
b. There is sufficient evidence to reject the null hypothesis.
Answer and Step-by-step explanation:
The parent function, f(x) = |x|, has no transformations applied to it.
When the number is with the x, it translates the graph horizontally. <->
When the number is outside of the absolute box, it translates the graph vertically. ^ V
When the number is multiply the absolute box on the outside with the x inside, it scales the graph.
So, first lets apply the translations of horizontal and vertical shift.
The graph translated 1 units downwards.
The graph translates 2 units to the right.
g(x) = |x - 2| - 1
We subtract the 2 because that makes it go right, while if we added it would go left.
Now for the scale.
The lines have a slope of , so that is what we are multiplying.
<u> g(x) = </u><u>|x - 2| - 1</u>
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<u>^That is your transformed function.</u>
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