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Montano1993 [528]
3 years ago
5

The average triglyceride level of adult males is 126 milligrams per deciliter. You want to determine whether the true mean male

triglyceride level is more than the average. You decide to pull a random sample of eight adult males and measure their triglyceride level. You find the sample mean is 127.9 milligrams per deciliter and the standard deviation is 4.12 milligrams per deciliter. What are the test statistic and the p-value?

Mathematics
1 answer:
crimeas [40]3 years ago
7 0

Answer: The test statistic is 1.30 and the p-value is 0.1167.

Step-by-step explanation:

A test statistic is a random variable which is calculated from sample data and also used in a hypothesis test. It can also be used to know whether to reject the null hypothesis.

The test statistic is 1.30 and the p-value is 0.1167. The solution is attached below

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The graph of y = StartAbsoluteValue x EndAbsoluteValue is transformed as shown in the graph below. Which equation represents the
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Using translation concepts, it is found that the equation of the transformed function is:

y = |x + 3| - 2.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

The absolute value function is given by y = |x|, and has vertex at (0,0). After the transformations, the function has vertex (-3, -2), which means that:

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Thus, the equation of the function is now given by:

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More can be learned about translation concepts at brainly.com/question/4521517

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The equation a<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class=
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Answer:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right)  = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Step-by-step explanation:

We are given the equation:

ax^2+bx+c=0

Which has roots α and β.

And we want to express (α + 1)(β + 1) in terms of <em>a</em>, <em>b</em>, and <em>c</em>.

From the quadratic formula, we know that the two solutions to our equation are:

\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Let <em>x</em>₁ = α and <em>x₂ </em>= β. Substitute:

\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)

Combine fractions:

\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)

Rewrite:

\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}

Multiply and group:

\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}

Difference of two squares:

\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}

Expand and simplify:

\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}

Distribute:

\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}

Cancel like terms:

\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}

Factor:

\displaystyle =\frac{4a(a+c-b)}{4a(a)}

Cancel. Hence:

\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Therefore:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}

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