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3 years ago

15

Explain how to use the regression calculator to make a reasonable prediction given a data table

2 answers:

solmaris [256]3 years ago

7 0

Answer:

Given data, first determine which is the independent variable, x, and which is the dependent variable, y. Enter the data pairs into the regression calculator. Substitute the value for one variable into the equation for the regression line produced by the calculator, and then predict the value of the other variable.

Step-by-step explanation:

Enter data into the regression calculator.
Determine the regression equation.
Substitute the correct value for x or y into the equation.
Simplify to find the value of the other variable.

riadik2000 [5.3K]3 years ago

3 0

Answer:

Given bivariate data, first determine which is the independent variable, x, and which is the dependent variable, y. Enter the data pairs into the regression calculator. Substitute the value for one variable into the equation for the regression line produced by the calculator, and then predict the value of the other variable.

Step-by-step explanation:

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gizmo_the_mogwai [7]

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$ .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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