Advanced Placement (AP), 21.12.2020 06:30 jenelllukus

There have been many studies recently concerning coffee drinking and cholesterol level. While it is known that several coffee-bean components can elevate blood cholesterol level, it is thought that a
new type of paper coffee filter may reduce the presence of some of these components in coffee.
The effect of the new filter on cholesterol level will be studied over a 10-week period using 300
nonsmokers who each drink 4 cups of caffeinated coffee per day. 150 participants are randomly
assigned to the experimental group, who will only drink coffee that has been made with the new
filter, with remaining participats assigned to the control group, who will only drink coffee that has
been made with the standard filter. Each participant's cholesterol level will be measured at the
beginning and at the ind of the study to determine if there was an increase in cholesterol level.
The table summarizes the result of the experiment:
New Filter
Old Filter
Increase in Cholesterol
62
84
No Increase in Cholesterol
88
66 -

a.
Does the data provide enough evidence to conclude that the new filter does a better job with
cholesterol?

b.
Estimate the difference in proportions for the effect on cholesterol for the two filters.

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Answer from: Quest

consider the following scenario, then describe how you might conduct a study using ansrs, convenience sample, and a voluntary response sample. you are interested in parents' opinion of the war in iraq. your population is parents of high school students. your question is, "do you believe that the war in

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

Answer from: Quest

See diagram.we use washers to get the area and we will be integrating with respect to xouter radius is the distance between y = e^x and y = -1 $r_{\text{out} } = e^x ) = e^x + 1$ inner radius of washer is the distance between y = 1 and y = -1 $r_{\text{in} } = 1 - (-1) = 2$ the area of one cross-section of a washer is $\begin{aligned} a(x) & = \pi(r_{\text{out} } )^2 - \pi(r_{\text{in} } )^2 \\ & = \pi(e^x + 1 )^2 - \pi(2)^2 \\ & = \pi \big[(e^x + 1 )^2 - 4 \big] \end{aligned}$ solid lies between x = 0 and x = 2. the volume is $\begin{aligned} v & = \int_0^2 a(x) dx \\ & = \int_0^2 \pi \big[(e^x + 1 )^2 - 4 \big] dx \end{aligned}$ if you need this expanded, then $\begin{aligned} v & = \int_0^2 \pi \left(e^{2x} + 2e^x + 1 - 4 \right) dx \\ & = \int_0^2 \pi \left(e^{2x} + 2e^x -3 \right) dx \end{aligned}$ the question doesn't say to evaluate, so we're done.

Answer from: Quest

Ithink its a im not sure

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