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4 years ago

How to solve the second equation

Mathematics

1 answer:

ivann1987 [24]4 years ago

4 0

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
y = &{{ -0.02}}x^2&{{ +1}}x&{{ +6}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{1^2}{2(-0.02)}~~,~~6-\cfrac{1^2}{4(-0.02)} \right)
\\\\\\
(\stackrel{\textit{how far ahead it went}}{25}~,~\stackrel{\textit{how high it went}}{18\frac{1}{2}})$

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
y = &{{ -0.01}}x^2&{{ +0.7}}x&{{ +6}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{0.7^2}{2(-0.01)}~~,~~6-\cfrac{0.7^2}{4(-0.01)} \right)
\\\\\\
(\stackrel{\textit{how far ahead it went}}{24\frac{1}{2}}~,~\stackrel{\textit{how high it went}}{18\frac{1}{2}})$

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3 years ago

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Answer:

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3 years ago

A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn

Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, $\overline x_1$ = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, $\overline x_2$ = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

$t_{\alpha /2}$ = $t_{0.005, \, 21}$ = 1.721

Therefore, we have;

$\left (6.59- 15.5 \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}$

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

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3 years ago

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saw5 [17]

Step-by-step explanation:

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First lets make a table for f(x)= x^2

Lets pick some numbers for x and find out y

x y=x^2

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-1 1

0 0

1 1

3 9

Now we use the same x values and make a table for g(x)

x $g(x)= \frac{1}{3} x^2$

-3 $g(x)= \frac{1}{3}(-3)^2=3$

-1 $g(x)= \frac{1}{3}(-1)^2=\frac{1}{3}$

0 0

1 $g(x)= \frac{1}{3}(1)^2=\frac{1}{3}$

3 $g(x)= \frac{1}{3}(3)^2=3$

Graph the above table

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Step-by-step explanation:

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