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Kruka [31]
2 years ago
5

Im bad at math and i need help​

Mathematics
2 answers:
Law Incorporation [45]2 years ago
7 0

Answer:

n•√n

here is an accurate representation

Colt1911 [192]2 years ago
3 0
I think it would just be n rad n
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Find the linear regression equation for the transformed data. x=1,2,3,4,5 y=13,19,37,91,253 log y=1.114,1.279,1.568,1.959,2,403
Talja [164]

Answer:

The answer is OPTION (D)log(y)=0.326x+0.687

<h2>Linear regression:</h2>

It is a linear model, e.g. a model that assumes a linear relationship between the input variables (x) and the single output variable (y)

The Linear regression equation for the transformed data:

We transform the predictor (x) values only. We transform the response (y) values only. We transform both the predictor (x) values and response (y) values.

(1, 13) 1.114

(2, 19) 1.279

(3, 37) 1.568

(4, 91) 1.959

(5, 253) 2.403

X            Y           Log(y)

1              13           1.114

2             19         1.740

3             37        2.543

4             91        3.381

5            253      4.226

Sum of X = 15

Sum of Y = 8.323

Mean X = 3

Mean Y = 1.6646

Sum of squares (SSX) = 10

Sum of products (SP) = 3.258

Regression Equation = ŷ = bX + a

b = SP/SSX = 3.26/10 = 0.3258

a = MY - bMX = 1.66 - (0.33*3) = 0.6872

ŷ = 0.3258X + 0.6872

The graph is plotted below:

The linear regression equation is  log(y)=0.326x+0.687

Learn more about Linear regression equation here:

brainly.com/question/3532703

#SPJ10

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2 years ago
Empirical rule of 139 cm and 193 cm
sesenic [268]
<span> 166cm would be your answer.</span>
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3 years ago
PLS HELP 30 ISH POINTS
Nina [5.8K]

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Simplify completely.<br> (2 + 3i) + (-4+i)
Aleksandr-060686 [28]

Answer:

-2+4i

Step-by-step explanation:

Simplify by combining the real and imaginary parts of each expression.

7 0
3 years ago
Read 2 more answers
Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

as desired. So one possible ODE would be

-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

(See "Euler-Cauchy equation" for more info)
6 0
3 years ago
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