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Lelechka [254]
3 years ago
14

In the theory of learning, the rate at which a subject is memorized is assumed to be proportional to the amount that is left to

be memorized. Assume that the rate at which material is forgotten is proportional to the amount memorized. Suppose M denotes the total amount of a subject to be memorized and A(t) is the amount memorized in time t > 0. Determine a differential equation for the amount A(t) when forgetfulness is taken into account. (Assume the constants of proportionality for the rate at which material is memorized and the rate at which material is forgotten are k1 > 0 and k2 > 0, respectively. Use A for A(t).)
dA/dt =
Mathematics
1 answer:
babymother [125]3 years ago
7 0

Answer:

dA/dt = k1(M-A) - k2(A)

Step-by-step explanation:

If M denote the total amount of the subject and A is the amount memorized, the amount that is left to be memorized is (M-A)

Then, we can write the sentence "the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized" as:

Rate Memorized = k1(M-A)

Where k1 is the constant of proportionality for the rate at which material is memorized.

At the same way, we can write the sentence: "the rate at which material is forgotten is proportional to the amount memorized" as:

Rate forgotten = k2(A)

Where k2 is the constant of proportionality for the rate at which material is forgotten.

Finally, the differential equation for the amount A(t) is equal to:

dA/dt = Rate Memorized - Rate Forgotten

dA/dt = k1(M-A)  - k2(A)

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Question 1) 17.4

Question 2) 1.7

Question 3) 2.8

Question 4) 8.6

Step-by-step explanation:

Mean absolute deviation is the average distance between the data points from the set to the mean point of the data set. It shows the variability in data or how much the data points are spread.

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The given problem has four sub-parts

Solution 1:

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            =  75

Mean absolute deviation= ( I 78-75 I + I 99-75 I+I 90-75 I + I 80-75 I +

                                             I 55-75 I + I 56-75 I + I 102-75 I + I 88- 75 I +

                                             I 60-75 I + I 42-75 I) /10

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Solution 2:

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Mean absolute deviation = ( I 10-10.5 I + I 13-10.5 I + I 7-10.5 I + I 12-10.5 I +

                                              I 9-10.5 I + I 8-10.5 I + I 12-10.5 I + I 10-10.5 I +

                                               I 11-10.5 I + I 3-10.5 I) /10

       =   17/10

       = 1.7

Solution 3:

Mean= (1+7+10+5+3+3+6+12+9+4)/10

            =  60/10

            =  6

Mean absolute deviation  = ( I 1-6 I + I 7-6 I + I 10-6 I + I 5-6 I + I 3-6 I +

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         = (28)/10

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Mean= (30+46+25+45+18+25+15+32+40+24)/10

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                                               I 40-30 I + I 24-30 I) /10

         = (86)/10

        =  8.6

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Option D

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Therefore, there is a common difference of 4 in each successive term to the previous term.

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7 0
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