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zlopas [31]
3 years ago
8

The value of 5x2 - 11x + 18 is when x = 2.4 and whenx = -3.

Mathematics
2 answers:
Virty [35]3 years ago
8 0

Answer:

20.4

96

Step-by-step explanation:

The value of 5x² - 11x + 18 is

5(2.4)²-11(2.4) +18 = 5(5.76) -26.4+18

= 28.8 -26.4+18 = 20.4 (when x = 2.4)

and 5(-3)² -11(-3) +18 = 45+33+18 = 96

(when x = -3)

ivann1987 [24]3 years ago
4 0

f(x)=5x^2-11x+18

\\ \sf\longmapsto f(2.4)

\\ \sf\longmapsto 5(2.4)^2-11(2.4)+18

\\ \sf\longmapsto 5(5.76)-26.4+18

\\ \sf\longmapsto 28.8-26.4+18

\\ \sf\longmapsto 2.4+18

\\ \sf\longmapsto 20.4

And

\\ \sf\longmapsto f(-3)

\\ \sf\longmapsto 5(-3)^2-11(-3)+18

\\ \sf\longmapsto 5(9)+33+18

\\ \sf\longmapsto 45+51

\\ \sf\longmapsto 96

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What expression can be used for estimating 868 divided by 28
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868 is about 870, and 28 is about 30, so your expression would be 870 divided by 30 which would be 29 and the actual answer is 31, so your estimate would be close to your actual answer.
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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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Since, the density in this case matches the density of thick milk and hence,therefore, the liquid in the bottle is milk.

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