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kow [346]
3 years ago
11

X

Mathematics
1 answer:
slava [35]3 years ago
6 0

Answer:

Equation 1 x=-16

Equation 2 m=-3

Step-by-step explanation:

Equation 1

3x-x=-24-6

2x=-32

x=-32/2

x=-16

Equation 2

-2m=16-10

-2m=6

m=6/-2

m=-3

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g Minimizing the sum of the squared deviations around the line is called: predictor regression. mean squared error technique. le
Alex787 [66]

Minimizing the sum of the squared deviations around the line is called Least square estimation.

It is given that the sum of squares is around the line.

Least squares estimations minimize the sum of squared deviations around the estimated regression function. It is between observed data, on the one hand, and their expected values on the other. This is called least squares estimation because it gives the least value for the sum of squared errors. Finding the best estimates of the coefficients is often called “fitting” the model to the data, or sometimes “learning” or “training” the model.

To learn more about regression visit: brainly.com/question/14563186

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3 0
1 year ago
I just need no. a please help me to prove this. ​
OleMash [197]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    and         cos A = cos B · cos C

scratchwork:

  A + B + C = π

               A = π - (B + C)

         cos A = cos [π - (B + C)]                              Apply cos

                    = - cos (B + C)                                    Simplify

                    = -(cos B · cos C - sin B · sin C)          Sum Identity

                    = sin B · sin C - cos B · cos C               Simplify

cos B · cos C = sin B · sin C - cos B · cos C               Substitution

2cos B · cos C = sin B · sin C                                        Addition

                     2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}                                     Division

                     2 = tan B · tan C

\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}

<u>Proof LHS → RHS</u>

Given:                              A + B + C = π

Subtraction:                     A = π - (B + C)

Apply tan:                  tan A = tan(π - (B + C))

Simplify:                               = - tan (B + C)

\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)

Substitution:                        = -(tan B + tan C)/(1 - 2)

Simplify:                               = -(tan B + tan C)/-1

                                            = tan B + tan C

LHS = RHS:   tan B + tan C = tan B + tan C  \checkmark

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2 years ago
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dimaraw [331]
I need help with a question in algebra 2. May you please help me? If that’s fine with you?
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MAXImum [283]
Needs more information.

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3 years ago
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Ann had 45 fliers to post around town. Last week, she posted 2/3 of them. She posted 1/3 of the remaining fliers. How many flier
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Answer:

48

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(1/4)16 = 4

48-32-4 = 12 still not posted

checking

32 + 4 + 12 = 48 CHECKS!

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2 years ago
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