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bezimeni [28]
3 years ago
12

Determine whether or not the given differential equation is separable, that is, whether it can be expressed in the form p(y) dy

= q(x) dx. dy/dx − 2y = 16 + 3xy + 24x
Mathematics
1 answer:
Flura [38]3 years ago
7 0

Answer:

Yes , it is separable

Step-by-step explanation:

We are given that DE

\frac{dy}{dx}-2y=16+3xy+24x

We have to find the the given DE is separable or not.

\frac{dy}{dx}-2y=16x+3xy+24x

\frac{dy}{dx}=16+3xy+24x+2y

\frac{dy}{dx}=16+24x+3xy+2y=8(2+3x)+y(3x+2)

\frac{dy}{dx}=(2+3x)(8+y)

\frac{dy}{8+y}=(2+3x)dx

When the DE is separable then it can be written as

p(y)dy=q(x)dx

Given DE is in Separable form.Therefore, it is separable.

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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