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myrzilka [38]
2 years ago
12

Find the general solution to the differential equation modeling how a person learns:

Mathematics
1 answer:
Luda [366]2 years ago
3 0

Here we want to solve differential equations, we will see that the general solution is:

y = A*e^{-t} + 100

We want to solve the differential equation:

\frac{dy}{dt}  = 100 - y

From this is pretty clear that y is an exponential function, with an exponent of -1*t.

We can write it generally as:

y = A*e^{-t} + B\\\\\frac{dy}{dt}  = -A*e^{-t}

Then if we set B = 100 we get:

y = A*e^{-t} + 100\\\\\frac{dy}{dt}  = -A*e^{-t} \\\\\frac{dy}{dt}  = -A*e^{-t} - 100 + 100 = -y + 100

So we just found the general form of the function.

Now we have two cases:

A) y(0) = 35

y(0) = A*e^{-0} + 100  = 35\\=  A + 100 = 35\\\\A = 35 - 100 = -65

In this case, the function is:

y = -65*e^{-t} + 100

B) y(0) = 125

125 = A*e^0 + 100\\\\\125 - 100 = A\\\\25 = A

In this case, the function is:

y = 25*e^{-t} + 100

Now we want to see which one of the two can represent how a person learns. Just look at the graph below:

The green line is the one for y(0) = 35, and the blue one is for y(0) = 125.

Notice that for small values of t, the blue function is really large, thus it can't really model how a person learns (is larger for smaller values of t than for larger values).

So y(0) = 35 represents better how a <em>person can learn</em> (but not exactly, because you can see that it eventually becomes almost constant, which is something that really does not happen) so the correct option is <u>D: none of the above.</u>

If you want to learn more, you can read:

brainly.com/question/353770

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A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

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