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pickupchik [31]
3 years ago
5

A percent is a fraction whose denominator is 100 1.sometimes 2.always 3. never

Mathematics
2 answers:
uranmaximum [27]3 years ago
8 0
2. always

Percents tell us how many parts a certain thing has out of a total of 100 parts; thus, the denominator is always 100.
Valentin [98]3 years ago
7 0
2 always because percentages are always considered a number over 100
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A=(x+6)(2x+2)  

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A=2x^2+14x+12
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-26 =3v +1 can i please get a answer
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Step-by-step explanation:

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Does the point (3, 83) satisfy the equation y = x2 – 6x – -92?
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A linear regression model is fitted to the data x y 37.0 65.0 36.4 67.2 35.8 70.3 34.3 71.9 33.7 73.8 32.1 75.7 31.5 77.9 with x
andrey2020 [161]

Answer:

b0= 144.59

b= -2.12

Se²= 1.02

99%CI E(Y/X=35): [68.78; 71.99]

Step-by-step explanation:

Hello!

I've arranged the given data:

X: 37.0, 36.4, 35.8, 34.3, 33.7, 32.1, 31.5

Y: 65.0, 67.2, 70.3, 71.9, 73.8, 75.7, 77.9

The equation of the linear regression model is:

Yi= β₀ + βXi + εi

Where

Yi is the dependent variable

Xi is the independent variable

εi represents the errors or residues

β₀ is the intercept of the line

β is the slope

The conditions to make a linear regression analysis are:

For each given value of X, there is a population of Y~N(μy;σy²)

Each value of Y is independent of the others.

The population variances of each population of Y are equal.

From these conditions the following characteristic is deduced:

εi~N(0;σ²)

The parameters of the regression are:

β₀, β, and σ²

If the conditions are met then you can estimate the regression line:

Yi= bo * bXi + ei.

And the point estimation of the parameters can be calculated using the formulas:

β₀ ⇒ b0= (∑y/n)-b(∑x/n)

β ⇒ b= [∑xy- ((∑x)(∑y))/n]/(∑x²-((∑x)²/n))

σ²⇒ Se²= 1/(n-2)*[∑y²-(∑y)²/n - b²(∑x²-(∑x)²/n)]

n= 7

∑y= 501.80

∑y²= 36097.88

∑x= 240.80

∑x²= 8310.44

∑xy= 17204.87

b0= 144.59

b= -2.12

Se²= 1.02

The estimated regression line is:

Yi= 144.59 -2.12Xi

You need to calculate a 99%CI E(Y/X=35), the formula is:

(b0 + bX0) ± t_{n-2;1-\alpha /2}*\sqrt{S_e^2(\frac{1}{n}+\frac{(X_0-X[bar])^2}{sumX^2-(\frac{(sumX)^2}{n} )} )}

(144.59 + (-2.12*35)) ± 4.032*\sqrt{1.02(\frac{1}{7}+\frac{(35-34.4)^2}{8310.44-(\frac{(240.80)^2}{7} )} )}

[68.78; 71.99]

With a 99% confidence level youd expect that the interval [68.78; 71.99] contains the true value of the average of Y when X= 35.

I hope it helps!

6 0
3 years ago
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