Advanced Placement (AP), 04.09.2020 23:01 damientran

Use the following terms to explain how an experiment works: hypothesis, operational definition, random selection, random assignment, control group, experimental group, independent variable, dependent variable, and confounding variable.

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Answer from: shay5878

An experiment starts with a question , which then turns into a hypothesis. every experiment has variables in it and each one has a title in the expiemernt which is the controlled group the experimental group which is what you are testing, you also have the dependent and independent variables. and the confounding variable

Answer from: akheel4818

100% on this in AP psych

Explanation:

An experiment has a numerous amount of factors and steps that go into it. To begin, The scientists have a research question and begin observing all the variables. Then, the scientists come up with a testable hypothesis; an educated guess/assumption. After finding this hypothesis they have to conduct background research, they will want to know what other researchers have already found, and what questions remain unresolved. After that, they have to figure out what kind of experiment would be best suited for this research along with what variables go into the experiment. On top of that, they will have to figure out what the operational definition will be in this experiment, the statement of procedures the researcher is going to use in order to measure a specific variable. The researcher now has to select the participants for the study, in this experiment they are going to use random selection since it is easy and they’re selected randomly. After the participants are selected, they are placed into two groups the control group, which is the group that stays the same. And the experimental group, which is the group being tested. In this experiment it will be random assignment, much like random selection, they are randomly sorted into the groups. Once the groups are sorted, The control group will be the dependent variable, staying the same throughout the experiment, and the experimental group will be the independent variable, the ones being tested. Obviously, for both groups there will be confounding variables, factors other than the independent variable that will cause a result.

Answer from: Quest

answerpopulace;

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)

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