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OLEGan [10]
1 year ago
8

Logistic regression is used when you want to: Answer choices A. Predict a dichotomous variable from continuous or dichotomous va

riables. B. Predict a continuous variable from dichotomous variables. C. Predict any categorical variable from several other categorical variables. D. Predict a continuous variable from dichotomous or continuous variables.Logistic regression is used when you want to: Answer choices A. Predict a dichotomous variable from continuous or dichotomous variables. B. Predict a continuous variable from dichotomous variables. C. Predict any categorical variable from several other categorical variables. D. Predict a continuous variable from dichotomous or continuous variables.
Mathematics
1 answer:
masya89 [10]1 year ago
6 0

Answer:

A. Predict a dichotomous variable from continuous or dichotomous variables.

Step-by-step explanation:

Logistic regression is used when you want to predict a dichotomous variable from continuous or dichotomous variables.

Mathematically, it is given by the expression;

Logistic regression y with x_{1}, x_{2}........x_{n}

Where;

y represents the dichotomous dependent variable.

x_{1}, x_{2}........x_{n} represents the predictable variables, which are categorical in nature such as alive or dead, win or lose, sick or healthy, pass or fail, etc.

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If the perimeter of a square is 5 feet, how many inches long is each side of the square?
marysya [2.9K]
Well since we know that the perimeter of a square is four times the length of one of its sides. We just have to divide 5 by 4 to get the length of one side:

5feet /4 sides = 1.25 feet

And to finish off, we have to convert feet to inches:

<u>1 foot      </u>= <u>12 inches</u>
1.25 feet    x inches

x inches = 12 inches x 1.25 feet ÷ 1 foot
x inches = 15 inches

Therefore, each side is 15 inches long.

Hope this helps!
3 0
2 years ago
Read 2 more answers
-4/5, -1/2, 0.25, -0.2 least to greatest
galina1969 [7]

Answer:

<em>-4/5,-1/2,-0.2,0.25</em>

Step-by-step explanation:

To order these, it would be best to convert them into decimals.

-4/5 is equal to -0.8

-1/2 is equal to -0.5

-0.8 is the least, because the further left on a number line, the less the number is.

-0.5 is next and then -0.2

Finally, 0.2 is the greatest

<em>-4/5,-1/2,-0.2,0.25</em>

<u>Hope this helps :-)</u>

3 0
2 years ago
Write the quadratic equation whose roots are
maks197457 [2]
The quadratic equation y = ax^2 +bx+c can be written in the form:
y = a(x-p)(x-q)
Where p and q are the roots of the equation.
Given p = 4, q = -2, a = 3, the quadratic will look like this:
y = 3(x-4)(x+2)
Finally, distribute and combine like terms to put it in standard form.
y = 3(x^2 - 4x+2x -8) \\  \\ y = 3(x^2 -2x-8) \\  \\ y = 3x^2 -6x -24
5 0
2 years ago
Find a12 of the sequence 1/4,7/12,11/12,5/4,
Ludmilka [50]

Answer:

Your ans is. a12 = 47/12

Step-by-step explanation:

First, you need to find if the series has a common ratio or a common difference between each term. Based from observation, there is a common difference of 1/3 so the series is an arithmetic series.

The solution for this problem goes like this

an=a1+(n-1)d

a12=1/n+(12-1)(1/3)

a12=47/12

Hope it helped you.. Please mark BRAINLIEST

Tysm

8 0
2 years ago
A certain cylindrical tank holds 20,000 gallons of water which can be drained from the bottom of the tank in 20 minutes the volu
vazorg [7]

Answer:

Impossible. t=30 minutes.

Step-by-step explanation:

We have the function:

v(t)=20000(1-\frac{t}{20})^2

Where v(t) represents the amount of gallons remaining after t minutes.

We want to find at what time t is the <em>instanteous </em>rate of change from the tank 1000 gallons per minute.

In order to determine the instantaneous rate of change, let's find the derivative of our function. So, take the derivative of both sides with respect to our time t:

\frac{d}{dt}[v(t)]=\frac{d}{dt}(20000(1-\frac{t}{20})^2]

On the right, let's move the coefficient outside:

v'(t)=20000\frac{d}{dt}[(1-\frac{t}{20})^2]

To differentiate, let's use the chain rule, which is:

u(v(x))=u'(v(x))\cdot v'(x)

Our u(x) is x² and v(x) is (1-t/20). So, u'(x) is 2x and v'(x) is -1/20. Therefore:

v'(t)=20000(2(1-\frac{t}{20})\cdot -\frac{1}{20})

Simplify:

v'(t)=-1000(2(1-\frac{t}{20}))

Simplify:

v'(t)=-1000(2-\frac{t}{10})

Distribute:

v'(t)=100t-2000

So, the instantaneous rate of change after time t is given by the above function.

To find when the instantaneous rate of change of the water is 1000 gallons per minute, substitute 100 for v'(t) and solve for t. So:

1000=100t-2000

Solve for t. Add 2000 to both sides:

3000=100t

Divide both sides by 100:

t=30

So, after 30 minutes, the instantaneous rate of change will be 1000 gallons per minute.

However, if we go back to our original function, our domain t is only defined between 0 and 20 minutes.

So, it is impossible for our instantaneous rate of change to ever reach 1000 gallons per minute.

6 0
2 years ago
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