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1 year ago

Solve -7g + 10 + 3g = -9 - 4g + 29 . How many solutions does this equation have?

Mathematics

1 answer:

Natasha2012 [34]1 year ago

3 0

Answer:

Solve for g:

10 - 4 g = 20 - 4 g

Subtract 10 - 4 g from both sides:

0 = 10

Hint: | Look for a false statement.

0 = 10 is trivially false:

Answer: (no solutions)

Step-by-step explanation:

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Answer:

N=27 participants

Step-by-step explanation:

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The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

When we conduct a multiple regression we want to know about the relationship between several independent or predictor variables and a dependent or criterion variable.

If we assume that we have $k$ independent variables and we have $j=1,\dots,j$ individuals, we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^n (y_j-\bar y)^2$

$SS_{regression}=SS_{model}=\sum_{j=1}^n (\hat y_{j}-\bar y)^2$

$SS_{error}=\sum_{j=1}^n (y_{j}-\hat y_j)^2$

And we have this property

$SST=SS_{regression}+SS_{error}$

The degrees of freedom for the model on this case is given by $df_{model}=df_{regression}=k=2$ where k =2 represent the number of variables.

The degrees of freedom for the error on this case is given by $df_{error}=N-k-1=24$ . Sinc we know k we can find N.

$N=24+k+1=24+2+1=27$

And the total degrees of freedom would be $df=N-1=27 -1 =26$

On this case the correct answer would be N=27 participants

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1 year ago

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1 year ago

The 300 people who were lined up outside the amusement park had all entered by 8:15 a.m. By 8:30, 550 more had entered. The numb

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Answer:

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Step-by-step explanation:

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1 year ago

What are the solutions to the quadratic equation (5y + 6)2 = 24? y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 End

Paladinen [302]

Answer:

there are two solutions:

a) $y=\frac{-6+2\sqrt{6} }{5}$ , and

b) $y=\frac{-6-2\sqrt{6} }{5}$

Step-by-step explanation:

In the equation: $(5y+6)^2=24$ , since a perfect square with the unknown "y" is isolated on the left of the equal sign, we start by applying the square root on both sides of the equality, and then on isolating the unknown: