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Lelechka [254]
2 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]2 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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Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

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Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
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