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Natalija [7]
3 years ago
12

Variable reinforcement schedules generally demonstrate greater resistance to extinction than fixed reinforcement schedules. T F

Physics
2 answers:
Marrrta [24]3 years ago
8 0

The answer to the given statement is true.

This is because variable reinforcement schedules are less predictable than fixed reinforcement schedules. There are two possible types of variable reinforcement schedules: variable-ratio or variable-interval.

Variable-ratio reinforcement schedule means that the reinforcement would only be given when the individual exhibits the desired behavior according to a number of responses – these numbers are unpredictable for the subject. Variable-interval uses the same approach, however, the unpredictability lies on the period of time from when the desired behavior appears to when the reinforcement would be given.

This unpredictability causes extinction to be harder for the reinforced behavior since the subject might think that the reward for their behavior might still be given.

makvit [3.9K]3 years ago
6 0

Answer:

TRUE!

Variable reinforcement schedules generally demonstrate greater resistance to extinction than fixed reinforcement schedules

Explanation:

edge 2020:D

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Why is figure 5 an unhelpful visualization tool for this data set? <br><br> Please help!
Paraphin [41]

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

8 0
3 years ago
What is -8,675,309.0 in scientific notation
Sindrei [870]
= 3.456 × 1011
(scientific notation)

= 3.456e11
(scientific e notation)

= 345.6 × 109
(engineering notation)
(billion; prefix giga- (G))

= 345600000000
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4 0
3 years ago
In 1958 a large earthquake in Alaska produced a tsunami. What was the approximate height of the tsunami?
lilavasa [31]
The approximate height of the tsunami in Alaska in 1958 is 1720ft
6 0
4 years ago
Read 2 more answers
If m = 2,000, p = 2. 25, and y= 6,000, what is velocity?
Karo-lina-s [1.5K]

The velocity is 6.75



The velocity in the equation stated above can be calculated as follows

m= 2,000
p= 2.25

y= 6000

velocity= 2.25 × 6000/ 2000
= 13500/2000
= 6.75


Hence the velocity is 6.75

Please see the link below for more information.
brainly.com/question/23547288?referrer=searchResults

6 0
2 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
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