Advanced Placement (AP), 06.10.2019 13:30 thestarlexyp32wpj

Kalama just got a job in an office. she never knows when her boss may come by her cubicle, so she makes sure to always be working, just in case. which schedule of reinforcement best describes her behavior?

Answers: 2

Show answers

Answers

Answer from: texdee43

Variable-Ratio Schedule

Explanation:

Schedule of reinforcement is a rule stating which instances of behavior will be reinforced. In some cases, a behavior might be reinforced every time it occurs while at other times it might not be reinforced at all.

Reinforcement schedules take place in both in natural learning situations and training situations.

Some schedules are better suited to certain types of training situations. In some cases, training might call for one schedule and then switch to another once the desired behavior has been taught.

A VARIABLE-RATIO schedule is a schedule of reinforcement where a response is reinforced after an unpredictable number of responses. This schedule creates a steady, high rate of responding.

NOTE: variable-ratio schedule is unpredictable.

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

Iwould say b but i’m not so sure

Another question on Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 10:00

Look at the front-wheel-drive vehicle shown in figure a-3. the engine in this vehicle is mounted so that the front of the engine faces the side of the vehicle, so the engine would be called a(an) a. in-line engine. b. v-type engine. c. horizontally opposed engine. d. transverse-mounted engine.

Answers: 1

Answer

Advanced Placement (AP), 23.06.2019 16:40

Which statement is true? a crt is a higher-level respiratory therapy certification. every crt is also an rrt a rrt is a certification that adds a specialization every rrt is also a crt

Answers: 1

Answer

Advanced Placement (AP), 25.06.2019 19:30

Ben can’t wrestle in this week’s match unless he loses two pounds. he decides to take a nap instead of exercising. this is an example of wasting time getting distracted allowing others to set goals

Answers: 1

Answer

Advanced Placement (AP), 26.06.2019 16:00

What are political action committees seeking when they give campaign contributions

Answers: 1

Answer

You know the right answer?

Kalama just got a job in an office. she never knows when her boss may come by her cubicle, so she ma...

Questions