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Vinvika [58]
3 years ago
10

Help me please and fast need it done by tonight

Mathematics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

3 \frac{1}{2}

Step-by-step explanation:

Given

\frac{\frac{1}{2} }{4} = \frac{x}{28} ( cross- multiply )

4x = \frac{1}{2} × 28 = 14 ( divide both sides by 4 )

x = \frac{14}{4} = \frac{7}{2} = 3 \frac{1}{2}

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After getting bored flying his kite, Tony decides to tie it to the ground and let it fly itself. The kite gets caught on a light
yaroslaw [1]
The answer will be -50

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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!

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Answer: what’s you’re question


Step-by-step explanation:


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