Advanced Placement (AP), 26.06.2019 21:40 tasnimabdallah971

Dr. williams conducted a survey in the mall, measuring the shopping habits of people ages 15, 25, 35, 45, and 55, with 10 people in each group. which of the following best describes her study? an experiment a cross-sectional study a case study a longitudinal study a double-blind study

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Answer from: barbaragomez27

A cross-sectional study

Explanation:

The best fit would be cross-sectional study, for one because it is observational, meaning there is no manipulation of variables like you would find in an experiment. Secondly, it is comparing populations within different groups, in this case age groups. Although, this can also be done through a longitudinal study, but as the name suggests, it is longitudinal so it is done over longer periods of time.

Answer from: Quest

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Answer from: Quest

The answer is (d) 7/2======we have $f(x) = \begin{cases} x & \text{ for } x \le 0 \\ x + 1 & \text{ for } x \ \textgreater \ 0 \end{cases}$ we also have an integral $\displaystyle\int_{-2}^1 x\, f(x)\; dx$ notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum: $\displaystyle\int_{-2}^1 x\, f(x)\; dx = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx$ notice that for the piecewise function f, the function has the equationf(x) = x when x ≤ 0. so for the integral $\int_{-2}^0 x\, f(x)\; dx$ , f(x) = x so it changes into $\begin{aligned} \int_{-2}^0 x\, f(x)\; dx & = \int_{-2}^0 x \cdot x\; dx \\ & = \int_{-2}^0 x^2\; dx \\ & = \left[\frac{x^3}{3}\right]_{-2}^0 & & \text{\footnotesize antidifferentiate} \\ & = 0 - \frac{(-2)^3}{3} \\ & = \frac{8}{3} \end{aligned}$ for the piecewise function f, the function has the equationf(x) = x+1 when x > 0. for the integral $\int_{0}^1 x\, f(x)\; dx$ , we have $\begin{aligned} \int_{0}^1 x\, f(x)\; dx & = \int_{0}^1 x(x+1)\; dx \\ & = \int_{0}^1 (x^2+x)\; dx \\ & = \left[ \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 \\ & = \frac{1}{3} + \frac{1}{2} - 0 = \frac{5}{6} \end{aligned}$ therefore, back to our splitted integral: $\begin{aligned} \displaystyle\int_{-2}^1 x\, f(x)\; dx & = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx \\ & = \frac{8}{3} + \frac{5}{6} \\ & = \frac{7}{2} \end{aligned}$
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